Exercise 7.5

Integrate the functions in Exercises 1 to 21.

${"type":"$","id":"1-0-0-0-0-0","code":"$1.\\,\\frac{x}{\\left(x+1\\right)\\left(x+2\\right)}$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1601109030876,"cs":"S8tA0V7J/w9i/kZ+zC45pg==","size":{"width":104,"height":25}}$

Solun:- Let f(x) = ${"id":"5-0-0","code":"$\\frac{x}{\\left(x+1\\right)\\left(x+2\\right)}$","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","ts":1601109074171,"cs":"lT88thRFgxukIQnjNAqmSg==","size":{"width":65,"height":20}}$

Integrate f(x):-

${"font":{"color":"#000000","size":12,"family":"Arial"},"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\frac{x}{\\left(x+1\\right)\\left(x+2\\right)}.dx$","type":"$","id":"2-0-0-0-0-0-0-0-0-0","ts":1601109092691,"cs":"nxqTBL9SAATKeot8qyO52w==","size":{"width":240,"height":28}}$

By partial Fraction:-

${"code":"$\\frac{x}{\\left(x+1\\right)\\left(x+2\\right)}=\\frac{A}{\\left(x+1\\right)}+\\frac{B}{\\left(x+2\\right)}$","type":"$","id":"5","font":{"family":"Arial","size":12,"color":"#000000"},"ts":1601109158961,"cs":"pESI5aHTmIupEUVJ9fiuOw==","size":{"width":229,"height":28}}$

⇒ x = A(x+2) + B(x+1) …....(1)

Put x = -2 in eq. 1:-

⇒ -2 = 0 + B(-1)

⇒ B = 2

Put x = -1 in eq. 1:-

⇒ -1 = A(1)

⇒ A = -1 then

${"font":{"size":12,"family":"Arial","color":"#000000"},"id":"4-0","type":"$","code":"$\\frac{x}{\\left(x+1\\right)\\left(x+2\\right)}=\\frac{-1}{\\left(x+1\\right)}+\\frac{2}{\\left(x+2\\right)}$","ts":1601109810858,"cs":"wWsQWQzbbWGAoWdDtZZeQA==","size":{"width":229,"height":28}}$

${"id":"2-0-0-0-0-0-0-0-1-0","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{-1}{\\left(x+1\\right)}+\\frac{2}{\\left(x+2\\right)}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{-1}{\\left(x+1\\right)}.dx+\\int_{}^{}\\frac{2}{\\left(x+2\\right)}.dx}\t\n\\end{align*}","ts":1601552684695,"cs":"myUrW4fzK9x9AmT5lMvh6A==","size":{"width":305,"height":80}}$

We know that:-

${"id":"3-0-0-0-0-0-0","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"rWfS7XhN2BxdW/dq8mtvAQ==","size":{"width":140,"height":18}}$

${"type":"$","id":"2-1-1-0-0-0-0-0-0-0-0-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\int_{}^{}f\\left(x\\right)dx=-\\log_{}\\left(x+1\\right)+2\\log_{}\\left(x+2\\right)+C$","ts":1601110134166,"cs":"j1M9Txw74NhwXVbEsp9dSQ==","size":{"width":302,"height":17}}$

We know that:-

n.log m = log mn

log m - log n = log (m/n)

${"id":"2-1-1-0-0-0-0-0-0-0-1-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\log_{}\\left(x+1\\right)+\\log_{}\\left(x+2\\right)^{2}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\frac{\\left(x+2\\right)^{2}}{\\left(x+1\\right)}+C}\t\n\\end{align*}","ts":1601111169922,"cs":"Pqw3Lmfge99RXm8jQ6nLfw==","size":{"width":304,"height":81}}$

${"font":{"family":"Arial","size":12,"color":"#000000"},"code":"$2.\\,\\frac{1}{x^{2}-9}$","id":"1-0-0-0-1-0","type":"$","ts":1601111318685,"cs":"p/OBnu7jDDywErxZZWS/LA==","size":{"width":60,"height":25}}$

Solun:- Let f(x) = ${"type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"5-0-1-0","code":"$\\frac{1}{x^{2}-9}$","ts":1601111332283,"cs":"/bJs/qhl8Xj93m2WMnX/+g==","size":{"width":32,"height":20}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{x^{2}-9}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\left(x-3\\right)\\left(x+3\\right)}.dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-0-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601111503222,"cs":"JyGJ40E+retuom3NzDBFqA==","size":{"width":236,"height":76}}$

By partial Fraction:-

${"type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"6-0-0","code":"$\\frac{1}{\\left(x-3\\right)\\left(x+3\\right)}=\\frac{A}{\\left(x-3\\right)}+\\frac{B}{\\left(x+3\\right)}$","ts":1601111543521,"cs":"lHd4m20kxMpfavUFxdWhnQ==","size":{"width":229,"height":28}}$

⇒ 1 = A(x + 3) + B(x - 3) …....(1)

Put x = 3 in eq. 1:-

⇒ 1 = A(6)

⇒ A = 1/6

Put x = -3 in eq. 1:-

⇒ 1 = B(-6)

⇒ B = -1/6 then

${"id":"6-1-0","code":"$\\frac{1}{\\left(x-3\\right)\\left(x+3\\right)}=\\frac{1}{6}.\\frac{1}{\\left(x-3\\right)}-\\frac{1}{6}.\\frac{1}{\\left(x+3\\right)}$","type":"$","font":{"family":"Arial","color":"#000000","size":12},"ts":1601464146903,"cs":"22tnseIhuPvbBdpGF1CPYQ==","size":{"width":280,"height":28}}$

${"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{6}.\\frac{1}{\\left(x-3\\right)}-\\frac{1}{6}.\\frac{1}{\\left(x+3\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{6}\\int_{}^{}\\frac{1}{\\left(x-3\\right)}.dx-\\frac{1}{6}\\int_{}^{}\\frac{1}{\\left(x+3\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-1-1-0","ts":1601464175859,"cs":"zrbY3tDb8ghXuFNAHTiVXQ==","size":{"width":340,"height":77}}$

We know that:-

${"id":"3-0-0-0-0-0-1","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"hsG+gdPENn6YsivEiPfYAw==","size":{"width":140,"height":18}}$

${"font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-1-1-0-0-0-0-0-0-0-0-1-0-0","type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\frac{1}{6}\\log_{}\\left(x-3\\right)-\\frac{1}{6}\\log_{}\\left(x+3\\right)+C$","ts":1601464205220,"cs":"qc0Z/Mc2LhKzJUTeb3Ht3Q==","size":{"width":302,"height":18}}$

We know that:-

log m - log n = log (m/n)

${"font":{"family":"Arial","color":"#000000","size":10},"type":"align*","id":"2-1-1-0-0-0-0-0-0-0-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{6}\\left[\\log_{}\\left(x-3\\right)-\\log_{}\\left(x+3\\right)\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{6}\\log_{}\\left|\\frac{x-3}{x+3}\\right|+C}\t\n\\end{align*}","ts":1601464257023,"cs":"KrOV23EqEKFabdWKRM0k1w==","size":{"width":306,"height":76}}$

${"type":"$","id":"1-0-0-0-1-1-0","font":{"color":"#000000","family":"Arial","size":12},"code":"$3.\\,\\frac{3x-1}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}$","ts":1601112384262,"cs":"oXPcFQoXrFo8BTdvZvnQrw==","size":{"width":141,"height":28}}$

Solun:- Let f(x) = ${"type":"$","id":"5-0-1-1-0","code":"$\\frac{3x-1}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}$","font":{"family":"Arial","color":"#000000","size":10},"ts":1601112397081,"cs":"P0TOgBdXeJX6vLqpp9ICeQ==","size":{"width":94,"height":21}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{3x-1}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}.dx}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-1-1-0","ts":1601112473544,"cs":"fnqIxrc03QVA0mh/Ksdr+g==","size":{"width":284,"height":36}}$

By partial Fraction:-

${"id":"6-0-1-0-0","font":{"family":"Arial","size":12,"color":"#000000"},"type":"$","code":"$\\frac{3x-1}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}=\\frac{A}{\\left(x-1\\right)}+\\frac{B}{\\left(x-2\\right)}+\\frac{C}{\\left(x-3\\right)}$","ts":1601112518588,"cs":"qBZygubrhuAzqo2dMprBMw==","size":{"width":340,"height":28}}$

⇒ 3x - 1 = A(x-2)(x-3) + B(x-3)(x-1) + A(x-1)(x-2) …....(1)

Put x = 1 in eq. 1:-

⇒ 2 = A(-1)(-2) + 0 + 0

⇒ A = 1

Put x = 2 in eq. 1:-

⇒ 5 = B(-1)(1)

⇒ B = -5

Put x = 3 in eq. 1:-

⇒ 8 = C(2)(1)

⇒ C = 4 then

${"id":"6-0-1-1-0","font":{"family":"Arial","color":"#000000","size":12},"code":"$\\frac{3x-1}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}=\\frac{1}{\\left(x-1\\right)}-\\frac{5}{\\left(x-2\\right)}+\\frac{4}{\\left(x-3\\right)}$","type":"$","ts":1601112925258,"cs":"IQ4cDgCxrqtbQBiTaTyCtQ==","size":{"width":340,"height":28}}$

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{1}{\\left(x-1\\right)}-\\frac{5}{\\left(x-2\\right)}+\\frac{4}{\\left(x-3\\right)}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\left(x-1\\right)}.dx-5\\int_{}^{}\\frac{1}{\\left(x-2\\right)}.dx+4\\int_{}^{}\\frac{1}{\\left(x-3\\right)}.dx}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-1-1-1-0","ts":1601552717030,"cs":"TsZiDbn9yvXkxoCJoskLXw==","size":{"width":445,"height":80}}$

We know that:-

${"id":"3-0-0-0-0-0-2","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"QBu8pnNIJogdpfCNA0KB8Q==","size":{"width":140,"height":18}}$

${"id":"2-1-1-0-0-0-0-0-0-0-0-1-0-1-0-0","code":"$\\int_{}^{}f\\left(x\\right)dx=\\log_{}\\left(x-1\\right)-5\\log_{}\\left(x-2\\right)+4\\log_{}\\left(x-3\\right)+C$","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","ts":1601114187247,"cs":"FT0QD698rWVHyngtAPoNDA==","size":{"width":390,"height":17}}$

We know that:-

n.log m = log mn

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-1-1-0-0-0-0-0-0-0-0-1-0-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left(x-1\\right)-\\log_{}\\left(x-2\\right)^{5}+\\log_{}\\left(x-3\\right)^{4}+C}\\\\\n\\end{align*}","ts":1601114353566,"cs":"rbQlZ1T9L57dL/v2G++UpA==","size":{"width":388,"height":36}}$

${"type":"$","code":"$4.\\,\\frac{x}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}$","id":"1-0-0-0-1-1-1-0","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1601114431410,"cs":"Wk7kMdA1gGDcRMFsPkeFhg==","size":{"width":141,"height":25}}$

Solun:- Let f(x) = ${"code":"$\\frac{x}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}$","id":"5-0-1-1-1-0","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1601114475462,"cs":"OccwmrvsCRu6e6TZHZgBDw==","size":{"width":94,"height":20}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}.dx}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-0-0-0-0-0-0-1-1-1-0","type":"align*","ts":1601114492207,"cs":"QqHZp6WswejOVrUOHzLHSg==","size":{"width":284,"height":36}}$

By partial Fraction:-

${"code":"$\\frac{x}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}=\\frac{A}{\\left(x-1\\right)}+\\frac{B}{\\left(x-2\\right)}+\\frac{C}{\\left(x-3\\right)}$","type":"$","font":{"family":"Arial","color":"#000000","size":12},"id":"6-0-1-0-1-0","ts":1601114519913,"cs":"LREjp4XYYC7tL9J5/imCxQ==","size":{"width":340,"height":28}}$

⇒ x = A(x-2)(x-3) + B(x-3)(x-1) + A(x-1)(x-2) …....(1)

Put x = 1 in eq. 1:-

⇒ 1 = A(-1)(-2) + 0 + 0

⇒ A = 1/2

Put x = 2 in eq. 1:-

⇒ 2 = B(-1)(1)

⇒ B = -2

Put x = 3 in eq. 1:-

⇒ 3 = C(2)(1)

⇒ C = 3/2 then

${"type":"$","code":"$\\frac{x}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}=\\frac{1}{2}\\frac{1}{\\left(x-1\\right)}-\\frac{2}{\\left(x-2\\right)}+\\frac{3}{2}\\frac{1}{\\left(x-3\\right)}$","id":"6-0-1-1-1-0","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1601116279445,"cs":"zweDvdziZOP0Hi3AOFJ2cQ==","size":{"width":372,"height":28}}$

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{1}{2}\\frac{1}{\\left(x-1\\right)}-\\frac{2}{\\left(x-2\\right)}+\\frac{3}{2}\\frac{1}{\\left(x-3\\right)}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x-1\\right)}.dx-2\\int_{}^{}\\frac{1}{\\left(x-2\\right)}.dx+\\frac{3}{2}\\int_{}^{}\\frac{1}{\\left(x-3\\right)}.dx}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-0-0-0-0-0-1-1-1-1-0","type":"align*","ts":1601552744191,"cs":"hEuV9s8kleQgN/ORNiMUsQ==","size":{"width":470,"height":80}}$

We know that:-

${"id":"3-0-0-0-0-0-3","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"jwA+NJHZ3ThYy7rRjQurvQ==","size":{"width":140,"height":18}}$

${"code":"$\\int_{}^{}f\\left(x\\right)dx=\\frac{1}{2}\\log_{}\\left(x-1\\right)-2\\log_{}\\left(x-2\\right)+\\frac{3}{2}\\log_{}\\left(x-3\\right)+C$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","id":"2-1-1-0-0-0-0-0-0-0-0-1-0-1-0-1-0-0","ts":1601116315255,"cs":"0me7cGP29gB3nCoFrDq1yg==","size":{"width":404,"height":18}}$

${"code":"$5.\\,\\frac{2x}{x^{2}+3x+2}$","id":"1-0-0-0-1-1-1-1-0","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1601116447020,"cs":"iKy+kJO3hXNmRzY6RFf0ig==","size":{"width":86,"height":25}}$

Solun:- Let f(x) = ${"id":"5-0-1-1-1-1-0","type":"$","code":"$\\frac{2x}{x^{2}+3x+2}$","font":{"family":"Arial","color":"#000000","size":10},"ts":1601116462283,"cs":"PBeEYzziL2vHtb6gLCEtvw==","size":{"width":52,"height":20}}$

Integrate f(x):-

${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{2x}{x^{2}+3x+2}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{2x}{x^{2}+2x+x+2}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{2x}{x\\left(x+2\\right)+1\\left(x+2\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{2x}{\\left(x+1\\right)\\left(x+2\\right)}.dx}\\\\\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-0","ts":1601116568789,"cs":"iQG847xcHK1UEkVbqO4S4Q==","size":{"width":272,"height":158}}$

By partial Fraction:-

${"code":"$\\frac{2x}{\\left(x+1\\right)\\left(x+2\\right)}=\\frac{A}{\\left(x+1\\right)}+\\frac{B}{\\left(x+2\\right)}$","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"id":"6-0-1-0-1-1-0-0","ts":1601116606283,"cs":"h61bZYwtjH+J/jFIqrEEmw==","size":{"width":229,"height":28}}$

⇒ 2x = A(x+2) + B(x+1)

Put x = -1 in eq. 1:-

⇒ -2 = A(1)

⇒ A = -2

Put x = -2 in eq. 1:-

⇒ -4 = B(-1)

⇒ B = 4 then

${"code":"$\\frac{2x}{\\left(x+1\\right)\\left(x+2\\right)}=\\frac{-2}{\\left(x+1\\right)}+\\frac{4}{\\left(x+2\\right)}$","type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"id":"6-0-1-0-1-1-1","ts":1601117035162,"cs":"HnNCR46DhxdUUgn75QF7Gg==","size":{"width":229,"height":28}}$

${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{-2}{\\left(x+1\\right)}+\\frac{4}{\\left(x+2\\right)}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={-2\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx+4\\int_{}^{}\\frac{1}{\\left(x+2\\right)}.dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-0","ts":1601552771070,"cs":"LU4rHvio2n62j+xiFf8JMg==","size":{"width":340,"height":80}}$

We know that:-

${"id":"3-0-0-0-0-0-4","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"aNbrAzoaB9UkYej90H2sCA==","size":{"width":140,"height":18}}$

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-2\\log_{}\\left(x+1\\right)+4\\log_{}\\left(x+2\\right)+C}\\\\\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-0-0-1-0-1-0-1-0-1-0","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601117795135,"cs":"bPhkrKsxjw2qSwiuTqnqZQ==","size":{"width":316,"height":36}}$

${"id":"1-0-0-0-1-1-1-1-1-0-0","type":"$","font":{"family":"Arial","color":"#000000","size":12},"code":"$6.\\,\\frac{1-x^{2}}{x\\left(1-2x\\right)}$","ts":1601117865801,"cs":"y0ECPGo1ucxhL/sXkmblcw==","size":{"width":80,"height":30}}$

Solun:- Let f(x) = ${"type":"$","code":"$\\frac{1-x^{2}}{x\\left(1-2x\\right)}$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"5-0-1-1-1-1-1-0-0","ts":1601117882252,"cs":"bClc9Tw4+vdIKfRUEleriA==","size":{"width":48,"height":24}}$

Integrate f(x):-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1-x^{2}}{x\\left(1-2x\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1-x^{2}}{x-2x^{2}}.dx}\\\\\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601119163481,"cs":"9oO9oB1CYbIE0zRV05mkAg==","size":{"width":204,"height":81}}$

The power of the numerator is equal to the denominator then

${"code":"$\\ldiv{x-2x^{2}}{1-x^{2}}$","type":"$","font":{"family":"Arial","color":"#000000","size":10},"id":"7-0","ts":1601119448702,"cs":"5Neh/5X50Oy8YwaXs8/56w==","size":{"width":108,"height":24}}$

By quotient rule:-

${"code":"\\begin{align*}\n{1-x^{2}}&={\\frac{1}{2}\\left(x-2x^{2}\\right)+\\left(1-\\frac{x}{2}\\right)}\\\\\n{1-x^{2}}&={\\frac{x}{2}\\left(1-2x\\right)+\\left(1-\\frac{x}{2}\\right)}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"8-0-0","type":"align*","ts":1601120267431,"cs":"RrDEHQCNPBqvyp3MMml5DA==","size":{"width":226,"height":65}}$

${"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\frac{1-x^{2}}{x\\left(1-2x\\right)}}&={\\frac{x\\left(1-2x\\right)}{2x\\left(1-2x\\right)}+\\frac{\\left(1-\\frac{x}{2}\\right)}{x\\left(1-2x\\right)}}\\\\\n{\\frac{1-x^{2}}{x\\left(1-2x\\right)}}&={\\frac{1}{2}+\\frac{1}{2}\\frac{\\left(2-x\\right)}{x\\left(1-2x\\right)}}\t\n\\end{align*}","id":"9-0-0","ts":1601120325223,"cs":"8wonWvRXX3sBy6/JM1idmg==","size":{"width":265,"height":84}}$

${"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-1-0-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{1}{2}+\\frac{1}{2}\\frac{\\left(2-x\\right)}{x\\left(1-2x\\right)}\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{2}.dx+\\int_{}^{}\\frac{1}{2}\\frac{\\left(2-x\\right)}{x\\left(1-2x\\right)}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}1.dx+\\frac{1}{2}\\int_{}^{}\\frac{\\left(2-x\\right)}{x\\left(1-2x\\right)}dx}\t\n\\end{align*}","ts":1601552794930,"cs":"5Z204sP0rif6JLqWM+A5DQ==","size":{"width":300,"height":124}}$

By partial Fraction:-

${"type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"id":"6-0-1-0-1-1-0-1-0-0-0-0","code":"$\\frac{\\left(2-x\\right)}{x\\left(1-2x\\right)}=\\frac{A}{x}+\\frac{B}{\\left(1-2x\\right)}$","ts":1601120538877,"cs":"J9RRudF7ia5pTpu3QkT0bA==","size":{"width":188,"height":32}}$

⇒ 2 - x = A(1 - 2x) + B(x)

Put x = 0 in eq. 1:-

⇒ 2 = A(1)

⇒ A = 2

Put x = 1/2 in eq. 1:-

⇒ 3/2 = B(1/2)

⇒ B = 3 then

${"code":"$\\frac{\\left(2-x\\right)}{x\\left(1-2x\\right)}=\\frac{2}{x}+\\frac{3}{\\left(1-2x\\right)}$","font":{"size":12,"family":"Arial","color":"#000000"},"type":"$","id":"6-0-1-0-1-1-0-1-0-1-0","ts":1601120626335,"cs":"xXRmMp0Dc/4iNSmWLzvwjA==","size":{"width":184,"height":32}}$

${"font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}1.dx+\\frac{1}{2}\\int_{}^{}\\left(\\frac{2}{x}+\\frac{3}{\\left(1-2x\\right)}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}1.dx+\\int_{}^{}\\frac{1}{x}.dx+\\frac{3}{2}\\int_{}^{}\\frac{1}{\\left(1-2x\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-0","type":"align*","ts":1601552821386,"cs":"dr5o0iZ1pajeVxlkliXfUQ==","size":{"width":377,"height":80}}$

We know that:-

${"id":"3-0-0-0-0-0-5-0","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"IVqvL6pXpNEZH2OaMw+oRg==","size":{"width":140,"height":18}}$

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-1-1-0-0-0-0-0-0-0-0-1-0-1-0-1-0-1-1-0-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x}{2}+\\log_{}x+\\frac{3}{2}\\frac{\\log_{}\\left(1-2x\\right)}{-2}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x}{2}+\\log_{}\\left|x\\right|-\\frac{3}{4}\\log_{}\\left(1-2x\\right)+C}\t\n\\end{align*}","ts":1601120903946,"cs":"tBxU/uuW9tnleZ1SIMZsOg==","size":{"width":310,"height":77}}$

${"code":"$7.\\,\\frac{x}{\\left(x^{2}+1\\right)\\left(x-1\\right)}$","type":"$","id":"1-0-0-0-1-1-1-1-1-1","font":{"color":"#000000","family":"Arial","size":12},"ts":1601121088377,"cs":"QDQ3MSqabQ+1sRolu9GVQg==","size":{"width":109,"height":25}}$

Solun:- Let f(x) = ${"type":"$","code":"$\\frac{x}{\\left(x^{2}+1\\right)\\left(x-1\\right)}$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"5-0-1-1-1-1-1-1","ts":1601121107778,"cs":"9EAEJuezcGHMNnLA0yzbTg==","size":{"width":69,"height":20}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x}{\\left(x^{2}+1\\right)\\left(x-1\\right)}.dx}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-1","ts":1601121139794,"cs":"083zCjxQcRJhfhpTdcBV6g==","size":{"width":241,"height":36}}$

By partial Fraction:-

${"type":"$","code":"$\\frac{x}{\\left(x^{2}+1\\right)\\left(x-1\\right)}=\\frac{Ax+B}{x^{2}+1}+\\frac{C}{\\left(x-1\\right)}$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"6-0-1-0-1-1-0-1-0-0-1","ts":1601121244492,"cs":"tSKgkD4gr5Ok17/qWIEEmA==","size":{"width":238,"height":29}}$

⇒ x = (Ax+B)(x - 1) + C(x2 + 1).......(1)

Put x = 1 in eq. 1:-

⇒ 1 = C(2)

⇒ C = 1/2

Put x = 0 in eq. 1:-

⇒ 0 = B(-1)+C

⇒ B = C = 1/2

Put x = -1 in eq. 1:-

⇒ -1 = (-A+B)(-2)+C(2)

⇒ -1 = (-A+½)(-2)+(½)(2)

⇒ -1 = 2A-1+1

⇒ -1 = 2A

⇒ A = -1/2 then

${"code":"\\begin{align*}\n{\\frac{x}{\\left(x^{2}+1\\right)\\left(x-1\\right)}}&={\\frac{\\frac{-1}{2}x+\\frac{1}{2}}{x^{2}+1}+\\frac{\\frac{1}{2}}{\\left(x-1\\right)}}\\\\\n{\\frac{x}{\\left(x^{2}+1\\right)\\left(x-1\\right)}}&={\\frac{1}{2}\\frac{\\left(-x+1\\right)}{x^{2}+1}+\\frac{1}{2}\\frac{1}{\\left(x-1\\right)}}\t\n\\end{align*}","id":"6-0-1-0-1-1-0-1-0-0-1","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"ts":1601122043385,"cs":"F9C3p3vVEAEJN6e873YI0Q==","size":{"width":304,"height":84}}$

${"id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-1","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{1}{2}\\frac{\\left(-x+1\\right)}{x^{2}+1}+\\frac{1}{2}\\frac{1}{\\left(x-1\\right)}\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\frac{\\left(-x+1\\right)}{x^{2}+1}.dx+\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x-1\\right)}.dx}\t\n\\end{align*}","ts":1601552869675,"cs":"rARye9w46mNF38fqusZhJQ==","size":{"width":353,"height":80}}$

${"font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[\\int_{}^{}\\frac{-x}{x^{2}+1}.dx+\\int_{}^{}\\frac{1}{x^{2}+1}.dx+\\int_{}^{}\\frac{1}{\\left(x-1\\right)}.dx\\right]}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[I_{1}+I_{2}+I_{3}\\right]}\t\n\\end{align*}","type":"align*","id":"10","ts":1601124614405,"cs":"pzfzAfObHbLbICTmtUGi3Q==","size":{"width":444,"height":78}}$

Calculate I1:-

${"font":{"family":"Arial","color":"#222222","size":12},"id":"11-0-0","code":"$I_{1}=\\int_{}^{}\\frac{-x}{x^{2}+1}.dx$","type":"$","ts":1601124690734,"cs":"pN7JfuQ4B8ajmS2VumB3qQ==","size":{"width":134,"height":24}}$

Let x2 + 1 = t

⇒ 2x.dx = dt

⇒ x.dx = (1/2).dt

${"code":"\\begin{align*}\n{I_{1}}&={-\\int_{}^{}\\frac{1}{t}.\\frac{dt}{2}}\\\\\n{I_{1}}&={\\frac{-1}{2}\\int_{}^{}\\frac{1}{t}.dt}\t\n\\end{align*}","id":"11-1-0","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","ts":1601124922661,"cs":"oIlSV/Ul2vrqQVXv93ilJg==","size":{"width":120,"height":76}}$

We know that:-

${"id":"3-0-0-0-0-0-6-0","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"XJjVKQnFlzqJJAGYAqc5qw==","size":{"width":140,"height":18}}$

${"type":"align*","id":"11-1-1-0","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{I_{1}}&={\\frac{-1}{2}\\log_{}\\left|t\\right|+C}\t\n\\end{align*}","ts":1601124998897,"cs":"Z7aRKX2vlhgbwgpHw9fwVA==","size":{"width":130,"height":32}}$

Put the value of t:-

${"code":"\\begin{align*}\n{I_{1}}&={\\frac{-1}{2}\\log_{}\\left|x^{2}+1\\right|+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"id":"11-1-1-1","ts":1601125091452,"cs":"qCQ5Ac4x/4e9TMGe/3hN/w==","size":{"width":169,"height":32}}$

Calculate I2:-

${"type":"$","code":"$I_{2}=\\int_{}^{}\\frac{1}{x^{2}+1}.dx$","font":{"family":"Arial","color":"#222222","size":12},"id":"11-0-1-0","ts":1601125301511,"cs":"VLLreIZRsnxegYIAla4VXA==","size":{"width":134,"height":25}}$

We know that:-

${"id":"3-0-0-0-0-0-6-1","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\int_{}^{}\\frac{1}{1+x^{2}}.dx=\\tan^{-1}x+c$","ts":1601125349321,"cs":"KP7hfN0RGBXJPU+9LTPVVA==","size":{"width":168,"height":20}}$

${"type":"$","id":"11-0-1-1","font":{"size":10,"color":"#222222","family":"Arial"},"code":"$I_{2}=\\tan^{-1}x+C$","ts":1601125397510,"cs":"3MH1xRV2+ZnM0xNJ4UFeUA==","size":{"width":116,"height":16}}$

Calculate I3:-

${"font":{"family":"Arial","color":"#000000","size":10},"id":"12-0","code":"$I_{3}=\\int_{}^{}\\frac{1}{\\left(x-1\\right)}.dx$","type":"$","ts":1601125469165,"cs":"Ey3QYjslf95jXfptByC+Cw==","size":{"width":108,"height":21}}$

${"id":"3-0-0-0-0-0-6","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"UpQy5SGY6fYofthERlkW3w==","size":{"width":140,"height":18}}$

${"id":"12-1","code":"$I_{3}=\\log_{}\\left|x-1\\right|+C$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601125497949,"cs":"KoXTahpUHwnQsD6NklN3Gg==","size":{"width":134,"height":16}}$

${"id":"2-1-1-0-0-0-0-0-0-0-0-1-0-1-0-1-0-1-1-1","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[\\frac{-1}{2}\\log_{}\\left|x^{2}+1\\right|+\\tan^{-1}x+\\log_{}\\left|x-1\\right|\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left|x-1\\right|-\\frac{1}{4}\\log_{}\\left|x^{2}+1\\right|+\\frac{1}{2}\\tan^{-1}x+C}\t\n\\end{align*}","ts":1601125621420,"cs":"ii4IFNX4dHhzmTdncutzfQ==","size":{"width":412,"height":77}}$

${"font":{"family":"Arial","size":12,"color":"#000000"},"type":"$","code":"$8.\\,\\frac{x}{\\left(x-1\\right)^{2}\\left(x+2\\right)}$","id":"1-0-0-0-1-1-1-1-1-0-1-0","ts":1601197718329,"cs":"YXpwG5xbrF/lrGePp2jdHQ==","size":{"width":109,"height":28}}$

Solun:- Let f(x) = ${"code":"$\\frac{x}{\\left(x-1\\right)^{2}\\left(x+2\\right)}$","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"id":"5-0-1-1-1-1-1-0-1-0","ts":1601197765953,"cs":"mKNM34Qo0Dgp4RXC8nWsuA==","size":{"width":69,"height":21}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x}{\\left(x-1\\right)^{2}\\left(x+2\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","ts":1601197847657,"cs":"CHCf7TPZv0RIFGp4GrWdFw==","size":{"width":241,"height":40}}$

By partial Fraction:-

${"code":"$\\frac{x}{\\left(x-1\\right)^{2}\\left(x+2\\right)}=\\frac{A}{\\left(x-1\\right)}+\\frac{B}{\\left(x-1\\right)^{2}}+\\frac{C}{\\left(x+2\\right)}$","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-0","font":{"size":12,"family":"Arial","color":"#000000"},"type":"$","ts":1601197908878,"cs":"wcEPs1rJ2iWPUSN6f9x5NQ==","size":{"width":313,"height":32}}$

⇒ x = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)2…....(1)

Put x = 1 in eq. 1:-

⇒ 1 = B(3)

⇒ B = 1/3

Put x = - 2 in eq. 1:-

⇒ - 2 = C(9)

⇒ C = -2/9

Put x = 0 in eq. 1:-

⇒ 0 = A(-2) + B(2) + C(1)

⇒ 0 = A(-2) +(2/3) + (-2/9)

⇒ 0 = A(-2) +(4/9)

⇒ A = 2/9 then

${"code":"$\\frac{x}{\\left(x-1\\right)^{2}\\left(x+2\\right)}=\\frac{2}{9}\\frac{1}{\\left(x-1\\right)}+\\frac{1}{3}\\frac{1}{\\left(x-1\\right)^{2}}+\\frac{-2}{9}\\frac{1}{\\left(x+2\\right)}$","id":"6-0-1-0-1-1-0-1-0-0-0-1-1","type":"$","font":{"color":"#000000","family":"Arial","size":12},"ts":1601199514373,"cs":"9pYBdssrykESmaqENsb6+w==","size":{"width":372,"height":32}}$

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{2}{9}\\frac{1}{\\left(x-1\\right)}+\\frac{1}{3}\\frac{1}{\\left(x-1\\right)^{2}}-\\frac{2}{9}\\frac{1}{\\left(x+2\\right)}\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2}{9}\\int_{}^{}\\frac{1}{\\left(x-1\\right)}.dx+\\frac{1}{3}\\int_{}^{}\\frac{1}{\\left(x-1\\right)^{2}}.dx-\\frac{2}{9}\\int_{}^{}\\frac{1}{\\left(x+2\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601553108808,"cs":"K/zy11VgD3+/81N8qF2Clg==","size":{"width":484,"height":92}}$

We know that:-

${"id":"3-0-0-0-0-0-5-1-0","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"khAt4r66g6orKBscrTiiow==","size":{"width":140,"height":18}}$

${"id":"13-0","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","ts":1601199722770,"cs":"6vje406qG5XWADtt5K+jCg==","size":{"width":130,"height":21}}$

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2}{9}\\log_{}\\left|x-1\\right|+\\frac{1}{3}\\frac{{\\left(x-1\\right)}^{-1}}{-1}-\\frac{2}{9}\\log_{}\\left(x+2\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2}{9}\\log_{}\\left|x-1\\right|-\\frac{1}{3\\left(x-1\\right)}-\\frac{2}{9}\\log_{}\\left(x+2\\right)+C}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"2-1-1-0-0-0-0-0-0-0-0-1-0-1-0-1-0-1-1-0-1-0","ts":1601469972561,"cs":"zMkcG4hHXIfnkCkH2nxKDA==","size":{"width":414,"height":81}}$

${"id":"1-0-0-0-1-1-1-1-1-0-1-1-0","type":"$","code":"$9.\\,\\frac{3x+5}{x^{3}-x^{2}-x+1}$","font":{"color":"#000000","size":12,"family":"Arial"},"ts":1601200007652,"cs":"7jDrgIfx50UT4JPilWzfew==","size":{"width":104,"height":26}}$

Solun:- Let f(x) = ${"id":"5-0-1-1-1-1-1-0-1-1-0","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\frac{3x+5}{x^{3}-x^{2}-x+1}$","ts":1601200033889,"cs":"hlO4nf4VdtoSGDoQf4CFkA==","size":{"width":66,"height":20}}$

Integrate f(x):-

${"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{3x+5}{x^{3}-x^{2}-x+1}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{3x+5}{\\left(x-1\\right)^{2}\\left(x+1\\right)}.dx}\t\n\\end{align*}","ts":1601200228672,"cs":"T1o6AuqwG48B8EoTeSNy9w==","size":{"width":244,"height":80}}$

By partial Fraction:-

${"font":{"size":12,"color":"#000000","family":"Arial"},"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-0","type":"$","code":"$\\frac{3x+5}{\\left(x-1\\right)^{2}\\left(x+1\\right)}=\\frac{A}{\\left(x-1\\right)}+\\frac{B}{\\left(x-1\\right)^{2}}+\\frac{C}{\\left(x+1\\right)}$","ts":1601200261538,"cs":"xE4bgTgojguqOnq/WKQVmQ==","size":{"width":313,"height":32}}$

⇒ 3x + 5 = A(x - 1)(x + 1) + B(x + 1) + C(x - 1)2…....(1)

Put x = 1 in eq. 1:-

⇒ 8 = 2B

⇒ B = 4

Put x = - 1 in eq. 1:-

⇒ 2 = 4C

⇒ C = 1/2

Put x = 0 in eq. 1:-

⇒ 5 = A(-1) + B(1) + C(1)

⇒ 5 = A(-1) + 4(1) +(1/2)

⇒ A = -1/2 then

${"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-1","font":{"size":12,"family":"Arial","color":"#000000"},"type":"$","code":"$\\frac{3x+5}{\\left(x-1\\right)^{2}\\left(x+1\\right)}=\\frac{-1}{2}\\frac{1}{\\left(x-1\\right)}+\\frac{4}{\\left(x-1\\right)^{2}}+\\frac{1}{2}\\frac{1}{\\left(x+1\\right)}$","ts":1601200677017,"cs":"7rbISFnHGOymeyxdWkANaw==","size":{"width":356,"height":32}}$

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{-1}{2}\\frac{1}{\\left(x-1\\right)}+\\frac{4}{\\left(x-1\\right)^{2}}+\\frac{1}{2}\\frac{1}{\\left(x+1\\right)}\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-1}{2}\\int_{}^{}\\frac{1}{\\left(x-1\\right)}.dx+4\\int_{}^{}\\frac{1}{\\left(x-1\\right)^{2}}.dx+\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601553135470,"cs":"srmXLoZfpgbLXhGSmabqqg==","size":{"width":488,"height":92}}$

We know that:-

${"id":"3-0-0-0-0-0-5-1-1","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"BhjRuWqs+gnYiwpxYwN5BA==","size":{"width":140,"height":18}}$

${"id":"13-1","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","ts":1601199722770,"cs":"aMky4ku3/WBYYEv9SH/4uA==","size":{"width":130,"height":21}}$

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-1}{2}\\log_{}\\left|x-1\\right|+\\frac{4{\\left(x-1\\right)}^{-1}}{-1}+\\frac{1}{2}\\log_{}\\left(x+1\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left(x+1\\right)-\\frac{1}{2}\\log_{}\\left|x-1\\right|-\\frac{4}{x-1}+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"14-0-0","ts":1601201025279,"cs":"HNT/Z9FoQN9eomE82Crnqw==","size":{"width":420,"height":80}}$

We know that

⇒ log m - log n = log (m/n)

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\frac{\\left|x+1\\right|}{\\left|x-1\\right|}-\\frac{4}{x-1}+C}\t\n\\end{align*}","id":"14-1","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601201122504,"cs":"iEOdo2zobeK8G2ODxY6jWA==","size":{"width":272,"height":37}}$

${"type":"$","code":"$10.\\,\\frac{2x-3}{\\left(x^{2}-1\\right)\\left(2x+3\\right)}$","font":{"color":"#000000","family":"Arial","size":12},"id":"1-0-0-0-1-1-1-1-1-0-1-1-1-0","ts":1601201291672,"cs":"ATOAEh0F2YP4HMnMFQF00g==","size":{"width":126,"height":28}}$

Solun:- Let f(x) = ${"type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\frac{2x-3}{\\left(x^{2}-1\\right)\\left(2x+3\\right)}$","id":"5-0-1-1-1-1-1-0-1-1-1-0","ts":1601201317239,"cs":"HvgXjbXDD2pmxKlawTEwLQ==","size":{"width":76,"height":22}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{2x-3}{\\left(x^{2}-1\\right)\\left(2x+3\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{2x-3}{\\left(x-1\\right)\\left(x+1\\right)\\left(2x+3\\right)}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601201364251,"cs":"IItRwSJKZZeXuNvNPpH8Zg==","size":{"width":292,"height":77}}$

By partial Fraction:-

${"code":"$\\frac{2x-3}{\\left(x-1\\right)\\left(x+1\\right)\\left(2x+3\\right)}=\\frac{A}{\\left(x-1\\right)}+\\frac{B}{\\left(x+1\\right)}+\\frac{C}{\\left(2x+3\\right)}$","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-0","font":{"size":12,"family":"Arial","color":"#000000"},"type":"$","ts":1601201419337,"cs":"ypneJH4iM2JnthK/ftk7Mw==","size":{"width":353,"height":28}}$

⇒ 2x - 3 = A(x + 1)(2x + 3) + B(x - 1)(2x + 3) + C(x - 1)(x + 1)

Put x = 1:-

⇒ -1 = A(2)(5)

⇒ A = -1/10

Put x = - 1:-

⇒ -5 = B(-2)(1)

⇒ B = 5/2

Put x = -3/2:-

⇒ -6 = C(-5/2)(-1/2)

⇒ C = -24/5 then

${"type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-1","code":"$\\frac{2x-3}{\\left(x-1\\right)\\left(x+1\\right)\\left(2x+3\\right)}=\\frac{-1}{10}\\frac{1}{\\left(x-1\\right)}+\\frac{5}{2}\\frac{1}{\\left(x+1\\right)}-\\frac{24}{5}\\frac{1}{\\left(2x+3\\right)}$","ts":1601202607007,"cs":"N6Da6IqYr3ksWMCeLWzMhw==","size":{"width":420,"height":28}}$

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{-1}{10}\\frac{1}{\\left(x-1\\right)}+\\frac{5}{2}\\frac{1}{\\left(x+1\\right)}-\\frac{24}{5}\\frac{1}{\\left(2x+3\\right)}\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-1}{10}\\int_{}^{}\\frac{1}{\\left(x-1\\right)}.dx+\\frac{5}{2}\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx-\\frac{24}{5}\\int_{}^{}\\frac{1}{\\left(2x+3\\right)}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-0","ts":1601553160410,"cs":"mfcjVV1+qCmRA6sIuczN1g==","size":{"width":505,"height":80}}$

We know that:-

${"id":"3-0-0-0-0-0-5-1-2","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"6m+td1WKXtLZcDeALOZong==","size":{"width":140,"height":18}}$

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-1}{10}\\log_{}\\left|x-1\\right|+\\frac{5}{2}\\log_{}\\left|x+1\\right|-\\frac{24}{5}\\frac{\\log_{}\\left(2x+3\\right)}{2}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{5}{2}\\log_{}\\left|x+1\\right|-\\frac{1}{10}\\log_{}\\left|x-1\\right|-\\frac{12}{5}\\log_{}\\left(2x+3\\right)+C}\t\n\\end{align*}","type":"align*","id":"14-0-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601202677035,"cs":"AKpcj3TFZXfKS8abeMFKOQ==","size":{"width":448,"height":77}}$

${"code":"$11.\\,\\frac{5x}{\\left(x+1\\right)\\left(x^{2}-4\\right)}$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-0","type":"$","font":{"family":"Arial","color":"#000000","size":12},"ts":1601202828333,"cs":"xnbsMmFbSd/Vhm9XRW5UFg==","size":{"width":120,"height":28}}$

Solun:- Let f(x) = ${"font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","id":"5-0-1-1-1-1-1-0-1-1-1-1-0","code":"$\\frac{5x}{\\left(x+1\\right)\\left(x^{2}-4\\right)}$","ts":1601202861686,"cs":"90Q88YZkmAD+HuQ86FsDvg==","size":{"width":69,"height":21}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{5x}{\\left(x+1\\right)\\left(x^{2}-4\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{5x}{\\left(x+1\\right)\\left(x-2\\right)\\left(x+2\\right)}.dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-0","type":"align*","ts":1601202945050,"cs":"nHE0DzavsdXwuKk0ZC9SIA==","size":{"width":284,"height":77}}$

By partial Fraction:-

${"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-0","code":"$\\frac{5x}{\\left(x+1\\right)\\left(x-2\\right)\\left(x+2\\right)}=\\frac{A}{\\left(x+1\\right)}+\\frac{B}{\\left(x-2\\right)}+\\frac{C}{\\left(x+2\\right)}$","type":"$","font":{"color":"#000000","family":"Arial","size":12},"ts":1601203071167,"cs":"LKROyRLV+QjWblBMvqGSwA==","size":{"width":340,"height":28}}$

⇒ 5x = A(x - 2)(x + 2) + B(x + 1)(x + 2) + C(x + 1)(x - 2)

Put x = -1:-

⇒ -5 = A(-3)(1)

⇒ A = 5/3

Put x = 2:-

⇒ 10 = B(3)(4)

⇒ B = 10/12 = 5/6

Put x = -2 in eq. 1:-

⇒ -10 = C(-1)(-4)

⇒ C = -10/4 = -5/2 then

${"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-1","code":"$\\frac{5x}{\\left(x+1\\right)\\left(x-2\\right)\\left(x+2\\right)}=\\frac{5}{3}\\frac{1}{\\left(x+1\\right)}+\\frac{5}{6}\\frac{1}{\\left(x-2\\right)}+\\frac{-5}{2}\\frac{1}{\\left(x+2\\right)}$","type":"$","font":{"color":"#000000","family":"Arial","size":12},"ts":1601203594723,"cs":"R7e4wKf1Dsv4HhgzDTInJw==","size":{"width":398,"height":28}}$

${"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{5}{3}\\frac{1}{\\left(x+1\\right)}+\\frac{5}{6}\\frac{1}{\\left(x-2\\right)}-\\frac{5}{2}\\frac{1}{\\left(x+2\\right)}\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{5}{3}\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx+\\frac{5}{6}\\int_{}^{}\\frac{1}{\\left(x-2\\right)}.dx-\\frac{5}{2}\\int_{}^{}\\frac{1}{\\left(x+2\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-0","ts":1601553193817,"cs":"7lh89vKCdMWio9RNPwrlQw==","size":{"width":477,"height":80}}$

We know that:-

${"id":"3-0-0-0-0-0-5-1-3-0","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"4osO9+RFrx4jwGFQlzgiAg==","size":{"width":140,"height":18}}$

${"id":"14-0-1-1-0-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{5}{3}\\log_{}\\left|x+1\\right|+\\frac{5}{6}\\log_{}\\left|x-2\\right|-\\frac{5}{2}\\log_{}\\left|x+2\\right|+C}\t\n\\end{align*}","type":"align*","ts":1601203821833,"cs":"FgjB0eSUrKgjeda8aqFhnA==","size":{"width":408,"height":36}}$

${"code":"$12.\\,\\frac{x^{3}+x+1}{x^{2}-1}$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":12},"type":"$","ts":1601203979768,"cs":"UPQui5AYGCaS24ktZXJEPQ==","size":{"width":89,"height":28}}$

Solun:- Let f(x) = ${"font":{"color":"#000000","size":10,"family":"Arial"},"code":"$\\frac{x^{3}+x+1}{x^{2}-1}$","type":"$","id":"5-0-1-1-1-1-1-0-1-1-1-1-1-0","ts":1601204007545,"cs":"Kmkaq3Qnk5ib83+YJVvh4g==","size":{"width":46,"height":22}}$

Integrate f(x):-

${"type":"align*","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{3}+x+1}{x^{2}-1}.dx}\t\n\\end{align*}","ts":1601204031966,"cs":"Ogm159SfyvrxU0QjbPx5pw==","size":{"width":209,"height":37}}$

The power of the numerator is greater than the denominator.

${"id":"7-1","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\ldiv{x^{2}-1}{x^{3}+x+1}$","ts":1601204198813,"cs":"iOLQ+7wew6MX1hyVjhnioA==","size":{"width":126,"height":24}}$

By quotient rule:-

${"type":"align*","id":"8-1","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{x^{3}+x+1}&={\\left(x^{2}-1\\right).\\left(x\\right)+\\left(2x+1\\right)}\\\\\n{\\frac{x^{3}+x+1}{x^{2}-1}}&={x+\\frac{2x+1}{x^{2}-1}}\t\n\\end{align*}","ts":1601204541805,"cs":"F3m5j9xOqSl11xJdut31wg==","size":{"width":260,"height":60}}$

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(x+\\frac{2x+1}{x^{2}-1}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}x.dx+\\int_{}^{}\\frac{2x+1}{x^{2}-1}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}x.dx+\\int_{}^{}\\frac{2x}{x^{2}-1}.dx+\\int_{}^{}\\frac{1}{x^{2}-1}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1601553297010,"cs":"izrzi3pDYkz63cDN0sEC9g==","size":{"width":366,"height":118}}$

Let x2 - 1 = t

Differentiate w.r.t. to t:-

2x.dx = dt

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}x.dx+\\int_{}^{}\\frac{dt}{t}+\\int_{}^{}\\frac{1}{x^{2}-1}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601204816843,"cs":"7CzjunetwU8RAjhpHGXKYA==","size":{"width":312,"height":36}}$

We know that:-

${"id":"3-0-0-0-0-0-5-1-4-0","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"YlpIRTmF7C+Siud+MMtKMA==","size":{"width":140,"height":18}}$

${"id":"3-0-0-0-0-0-5-1-4-1","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","font":{"family":"Arial","color":"#000000","size":10},"type":"$","ts":1601204871150,"cs":"ZsOjM1/GbvhbQvGgIITJIg==","size":{"width":130,"height":21}}$

${"font":{"family":"Arial","color":"#000000","size":10},"id":"3-0-0-0-0-0-5-1-4-2","code":"$\\int_{}^{}\\frac{1}{x^{2}-a^{2}}.dx=\\frac{1}{2a}\\log_{}\\left|\\frac{x-a}{x+a}\\right|+c$","type":"$","ts":1601204978921,"cs":"ktwINf2Or6wQTEz73LEKZw==","size":{"width":202,"height":20}}$

${"font":{"family":"Arial","size":10,"color":"#000000"},"id":"14-0-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}}{2}+\\log_{}\\left|t\\right|+\\frac{1}{2}\\log_{}\\left|\\frac{x-1}{x+1}\\right|+C}\t\n\\end{align*}","type":"align*","ts":1601205051723,"cs":"sXb2/UO84I83UXmnOph97g==","size":{"width":310,"height":37}}$

Put the value of t:-

${"font":{"family":"Arial","size":10,"color":"#000000"},"id":"14-0-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}}{2}+\\log_{}\\left|x^{2}-1\\right|+\\frac{1}{2}\\log_{}\\left|\\frac{x-1}{x+1}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}}{2}+\\log_{}\\left|\\left(x-1\\right)\\left(x+1\\right)\\right|+\\frac{1}{2}\\log_{}\\left|\\frac{x-1}{x+1}\\right|+C}\t\n\\end{align*}","type":"align*","ts":1601205777662,"cs":"xAURUkI2TiJaJn/aKqSAWA==","size":{"width":402,"height":80}}$

We know that:-

log (m.n) = log m + log n

log (m/n) = log m - log n

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}}{2}+\\log_{}\\left|x-1\\right|+\\log_{}\\left|x+1\\right|+\\frac{1}{2}\\log_{}\\left|x-1\\right|-\\frac{1}{2}\\log_{}\\left|x+1\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}}{2}+\\frac{3}{2}\\log_{}\\left|x-1\\right|+\\frac{1}{2}\\log_{}\\left|x+1\\right|+C}\t\n\\end{align*}","font":{"family":"Arial","size":8,"color":"#000000"},"id":"14-0-1-1-1-1-1","type":"align*","ts":1601205730266,"cs":"c7oSUXevMiII2bqytPdc/w==","size":{"width":448,"height":68}}$

${"code":"$13.\\,\\frac{2}{\\left(1-x\\right)\\left(1+x^{2}\\right)}$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1601205904041,"cs":"kJ/u1s8MMeArGI6moBo8kw==","size":{"width":120,"height":28}}$

Solun:- Let f(x) = ${"font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","code":"$\\frac{2}{\\left(1-x\\right)\\left(1+x^{2}\\right)}$","id":"5-0-1-1-1-1-1-0-1-1-1-1-1-1-0","ts":1601205922316,"cs":"V4nGQIiz5GhGUuYoAKGmEw==","size":{"width":69,"height":21}}$

Integrate f(x):-

${"type":"align*","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{2}{\\left(1-x\\right)\\left(1+x^{2}\\right)}.dx}\t\n\\end{align*}","ts":1601205939177,"cs":"BSEvAwIXNxtr9UPKXEt1Tg==","size":{"width":241,"height":36}}$

By partial Fraction:-

${"font":{"family":"Arial","size":12,"color":"#000000"},"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-0","type":"$","code":"$\\frac{2}{\\left(1-x\\right)\\left(1+x^{2}\\right)}=\\frac{A}{\\left(1-x\\right)}+\\frac{Bx+C}{\\left(1+x^{2}\\right)}$","ts":1601206652962,"cs":"YzsIEZuyalH/GwRDkR2u6g==","size":{"width":241,"height":29}}$

⇒ 2 = A(1 + x2) + (Bx+C)(1 - x)

Put x = 1:-

⇒ 2 = A(2)

⇒ A = 1

Put x = 0:-

⇒ 2 = A(1) + C(1)

⇒ 2 =(1) + C(1)

⇒ C = 1

Put x = -1:-

⇒ 2 = A(2) + (-B+C)(2)

⇒ 1 = A - B + C

⇒ 1 = 1 - B + 1

⇒ B = 1 then

${"type":"$","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-1","code":"$\\frac{2}{\\left(1-x\\right)\\left(1+x^{2}\\right)}=\\frac{1}{\\left(1-x\\right)}+\\frac{x}{\\left(1+x^{2}\\right)}+\\frac{1}{\\left(1+x^{2}\\right)}$","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1601207006921,"cs":"luqVL7wJ0CehrkIOtoF/8Q==","size":{"width":318,"height":28}}$

${"font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{1}{\\left(1-x\\right)}+\\frac{x}{\\left(1+x^{2}\\right)}+\\frac{1}{\\left(1+x^{2}\\right)}\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\left(1-x\\right)}.dx+\\int_{}^{}\\frac{x}{\\left(1+x^{2}\\right)}.dx+\\int_{}^{}\\frac{1}{\\left(1+x^{2}\\right)}.dx}\t\n\\end{align*}","type":"align*","ts":1601553324903,"cs":"dJWbXdr55WNPxSD0Gq03wg==","size":{"width":437,"height":80}}$

Let 1 + x2 = t

Differentiate w.r.t. to t:-

2x.dx = dt

x.dx = (1/2).dt

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{-1}{\\left(x-1\\right)}.dx+\\frac{1}{2}\\int_{}^{}\\frac{dt}{t}+\\int_{}^{}\\frac{1}{\\left(1+x^{2}\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-1","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601207392183,"cs":"LreL4U13mVkp/CRbqZtgiA==","size":{"width":389,"height":36}}$

We know that:-

${"id":"3-0-0-0-0-0-5-1-3-1-0","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"nlF/HQq7t2ZH4mhcb2+oAA==","size":{"width":140,"height":18}}$

${"type":"$","id":"15-0","code":"$\\int_{}^{}\\frac{1}{1+x^{2}}.dx=\\tan^{-1}x+c$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601206530493,"cs":"A43Wb5jU+Ax5sPz1teI50A==","size":{"width":168,"height":20}}$

${"font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\log_{}\\left|x-1\\right|+\\frac{1}{2}\\log_{}\\left|t\\right|+\\tan^{-1}x+C}\t\n\\end{align*}","id":"14-0-1-1-0-1-0-0","ts":1601547484176,"cs":"iX/UIr1Yw+RF39Z0A8TM1g==","size":{"width":344,"height":36}}$

Put the value of t:-

${"id":"14-0-1-1-0-1-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\log_{}\\left|x-1\\right|+\\frac{1}{2}\\log_{}\\left|1+x^{2}\\right|+\\tan^{-1}x+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1601547511225,"cs":"L+rF87H6+mhK1DN16gsNIg==","size":{"width":382,"height":36}}$

${"type":"$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":12},"code":"$14.\\,\\frac{3x-1}{\\left(x+2\\right)^{2}}$","ts":1601209038963,"cs":"gfCqcaLo7lQYPJ/N7Kt04A==","size":{"width":81,"height":32}}$

Solun:- Let f(x) = ${"code":"$\\frac{3x-1}{\\left(x+2\\right)^{2}}$","font":{"color":"#000000","size":10,"family":"Arial"},"id":"5-0-1-1-1-1-1-0-1-1-1-1-1-1-1-0","type":"$","ts":1601209053594,"cs":"wdR8gY+jBy6DIVqFP1R1XQ==","size":{"width":40,"height":24}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{3x-1}{\\left(x+2\\right)^{2}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1601209068520,"cs":"Se+d4Tp9taaKHOIIk7Pi9g==","size":{"width":193,"height":40}}$

By partial Fraction:-

${"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-0","font":{"size":12,"color":"#000000","family":"Arial"},"type":"$","code":"$\\frac{3x-1}{\\left(x+2\\right)^{2}}=\\frac{A}{\\left(x+2\\right)}+\\frac{B}{\\left(x+2\\right)^{2}}$","ts":1601209099803,"cs":"lG06m9LwAKPKNo5nRekaVA==","size":{"width":204,"height":32}}$

⇒ 3x - 1 = A(x + 2) + B

Put x = - 2:-

⇒ - 7 = B

⇒ B = -7

Put x = 0:-

⇒ -1 = A(2) + B

⇒ -1 = A(2) - 7

⇒ 2A = 6

⇒ A = 3 then

${"type":"$","font":{"family":"Arial","color":"#000000","size":12},"code":"$\\frac{3x-1}{\\left(x+2\\right)^{2}}=\\frac{3}{\\left(x+2\\right)}-\\frac{7}{\\left(x+2\\right)^{2}}$","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-1","ts":1601209237383,"cs":"gRmbaydHr/pJBwvw1zGPdw==","size":{"width":204,"height":32}}$

We know that:-

${"type":"align*","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{3}{\\left(x+2\\right)}-\\frac{7}{\\left(x+2\\right)^{2}}\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={3\\int_{}^{}\\frac{1}{\\left(x+2\\right)}.dx-7\\int_{}^{}\\frac{1}{\\left(x+2\\right)^{2}}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601553354577,"cs":"wMq1tnC0rvaAjjXopmU4/g==","size":{"width":333,"height":92}}$

${"id":"3-0-0-0-0-0-5-1-3-1-1","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"Bz6ylgJVjHHt1OSEeuDFYw==","size":{"width":140,"height":18}}$

${"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","id":"15-1-0","type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601209293120,"cs":"sXwjKDVPqjT0N0DdBNYPFw==","size":{"width":130,"height":21}}$

${"id":"14-0-1-1-0-1-0-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={3\\log_{}\\left|x+2\\right|-\\frac{7\\left(x+2\\right)^{-1}}{-1}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={3\\log_{}\\left|x+2\\right|+\\frac{7}{\\left(x+2\\right)}+C}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1601209392879,"cs":"D0SGsgj4vIY/M+wuCvknhQ==","size":{"width":296,"height":81}}$

${"font":{"family":"Arial","size":12,"color":"#000000"},"code":"$15.\\,\\frac{1}{x^{4}-1}$","type":"$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1","ts":1601286770720,"cs":"8IZeqLREUGGXVBwNq9u8rw==","size":{"width":70,"height":26}}$

Solun:- Let f(x) = ${"code":"$\\frac{1}{x^{4}-1}$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","id":"5-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-0","ts":1601286813225,"cs":"y/lziTlSYWeFkJ9390o8DA==","size":{"width":32,"height":20}}$

Integrate f(x):-

${"type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{x^{4}-1}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\left(x^{2}-1\\right)\\left(x^{2}+1\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\left(x-1\\right)\\left(x+1\\right)\\left(x^{2}+1\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-0","ts":1601287007323,"cs":"ZWwB6VSkioUbNVogxPelQg==","size":{"width":290,"height":118}}$

By partial Fraction:-

${"code":"$\\frac{1}{\\left(x-1\\right)\\left(x+1\\right)\\left(x^{2}+1\\right)}=\\frac{A}{\\left(x-1\\right)}+\\frac{B}{\\left(x+1\\right)}+\\frac{Cx+D}{\\left(x^{2}+1\\right)}$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-0","type":"$","ts":1601288602076,"cs":"RTyXmKWIuABdQow/tOrr9g==","size":{"width":350,"height":29}}$

⇒ 1 = A(x + 1)(x2 + 1) + B(x-1)(x2 + 1) + (Cx+D)(x - 1)(x + 1)

Put x = 1:-

⇒ 1 = A(2)(2)

⇒ A = 1/4

Put x = - 1:-

⇒ 1 = B(-2)(2)

⇒ B = -1/4

Put x = 0:-

⇒ 1 = A(1)(1) + B(-1)(1) + D(-1)(1)

⇒ 1 = (1/4) +(-1)(-1/4) + D(-1)(1)

⇒ 1 = (2/4) + D(-1)(1)

⇒ D = -1/2

Put x = 2:-

⇒ 1 = A(3)(5) + (-1/4)(1)(5) + (2C-1/2)(1)(3)

⇒ 1 = (15/4) +(-5/4) + 6C - 3/2

⇒ C = 0 then

${"code":"\\begin{align*}\n{\\frac{1}{\\left(x-1\\right)\\left(x+1\\right)\\left(x^{2}+1\\right)}}&={\\frac{1}{4}\\frac{1}{\\left(x-1\\right)}-\\frac{1}{4}\\frac{1}{\\left(x+1\\right)}+\\frac{0.x-\\frac{1}{2}}{\\left(x^{2}+1\\right)}}\\\\\n{\\frac{1}{\\left(x-1\\right)\\left(x+1\\right)\\left(x^{2}+1\\right)}}&={\\frac{1}{4}\\frac{1}{\\left(x-1\\right)}-\\frac{1}{4}\\frac{1}{\\left(x+1\\right)}-\\frac{1}{2\\left(x^{2}+1\\right)}}\t\n\\end{align*}","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-1","font":{"size":8,"color":"#000000","family":"Arial"},"type":"align*","ts":1601548970533,"cs":"9cE9xBac1bafKLggyaZI/w==","size":{"width":368,"height":70}}$

${"font":{"family":"Arial","color":"#000000","size":8},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{1}{4}\\frac{1}{\\left(x-1\\right)}-\\frac{1}{4}\\frac{1}{\\left(x+1\\right)}-\\frac{1}{2\\left(x^{2}+1\\right)}\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\int_{}^{}\\frac{1}{\\left(x-1\\right)}.dx-\\frac{1}{4}\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx-\\frac{1}{2}\\int_{}^{}\\frac{1}{\\left(x^{2}+1\\right)}.dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-1-1-0","ts":1601553394913,"cs":"Gb+LM3QoaHYcifsKWZUTpQ==","size":{"width":414,"height":68}}$

We know that:-

${"id":"3-0-0-0-0-0-5-1-3-1-2","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"XFqdPLE0kIX+QH88FNodNQ==","size":{"width":140,"height":18}}$

${"font":{"color":"#000000","family":"Arial","size":10},"code":"$\\int_{}^{}\\frac{1}{1+x^{2}}.dx=\\tan^{-1}x+c$","id":"15-1-1-0","type":"$","ts":1601288139333,"cs":"IJo1pFr1zzKBUukRGjRqXQ==","size":{"width":168,"height":20}}$

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\log_{}\\left|x-1\\right|-\\frac{1}{4}\\log_{}\\left|x+1\\right|-\\frac{1}{2}\\tan^{-1}x+C}\\\\\n\\end{align*}","id":"14-0-1-1-0-1-0-1-1-0","font":{"color":"#000000","family":"Arial","size":8},"ts":1601296812599,"cs":"dlJzYFduybviYjYc+XH4cg==","size":{"width":333,"height":30}}$

Put the value of t:-

${"font":{"size":8,"family":"Arial","color":"#000000"},"id":"14-0-1-1-0-1-0-1-1-1","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\log_{}\\left|x-1\\right|-\\frac{1}{4}\\log_{}\\left|x+1\\right|-\\frac{1}{2}\\tan^{-1}x+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\log_{}\\left|\\frac{x-1}{x+1}\\right|-\\frac{1}{2}\\tan^{-1}x+C}\t\n\\end{align*}","ts":1601298886859,"cs":"pgDP9yXV+1lXeHYChBINEg==","size":{"width":333,"height":64}}$

${"type":"$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-2-0","code":"$16.\\,\\frac{1}{x\\left(x^{n}+1\\right)}$","font":{"color":"#000000","size":12,"family":"Arial"},"ts":1601291984018,"cs":"1oLUxwOBUzh+ii2L0TsbCw==","size":{"width":90,"height":28}}$

Solun:- Let f(x) = ${"type":"$","id":"5-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-0","code":"$\\frac{1}{x\\left(x^{n}+1\\right)}$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601292019571,"cs":"BLoaEs0T8P61sK7KCqqf1g==","size":{"width":48,"height":21}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{x\\left(x^{n}+1\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{n-1}}{x^{n}\\left(x^{n}+1\\right)}.dx}\\\\\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-0","ts":1601292399481,"cs":"SicBgbALgPnygpSHRnaDCQ==","size":{"width":212,"height":80}}$

Let xn + 1 = t

Differentiate w.r.t. to t:-

⇒ nxn-1dx = dt

⇒ xn-1dx = (1/n).dt

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{t\\left(t-1\\right)}.\\frac{dt}{n}}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{n}\\int_{}^{}\\frac{1}{t\\left(t-1\\right)}.dt}\\\\\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-1-0","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601292477500,"cs":"MLXuV3yVhRSObUa3XJGKWA==","size":{"width":205,"height":77}}$

By partial Fraction:-

${"code":"$\\frac{1}{t\\left(t-1\\right)}=\\frac{A}{t}+\\frac{B}{\\left(t-1\\right)}$","type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-0","ts":1601292550316,"cs":"rIqgdow3rLnMnD8nLbcfdA==","size":{"width":164,"height":28}}$

⇒ 1 = A(t - 1) + B(t)

Put t = 0:-

⇒ 1 = A(-1)

⇒ A = -1

Put t = 1:-

⇒ 1 = B

⇒ B = 1 then

${"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-1","type":"$","code":"$\\frac{1}{t\\left(t-1\\right)}=\\frac{-1}{t}+\\frac{1}{\\left(t-1\\right)}$","font":{"color":"#000000","family":"Arial","size":12},"ts":1601292919930,"cs":"dSIbJOmxeEL4Pq4LhB49DQ==","size":{"width":172,"height":28}}$

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{n}\\int_{}^{}\\left[\\frac{-1}{t}+\\frac{1}{\\left(t-1\\right)}\\right].dt}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{n}\\left[-\\int_{}^{}\\frac{1}{t}.dx+\\int_{}^{}\\frac{1}{\\left(t-1\\right)}.dt\\right]}\t\n\\end{align*}","font":{"family":"Arial","size":8,"color":"#000000"},"id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-1-1-2-0","ts":1601553419799,"cs":"ikOuuE7yxiY+xGy52uLofQ==","size":{"width":262,"height":68}}$

We know that:-

${"id":"3-0-0-0-0-0-5-1-3-1-3","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"GYwaJpH5aqcvjciEmXdspA==","size":{"width":140,"height":18}}$

${"type":"align*","id":"14-0-1-1-0-1-0-1-1-2-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{n}\\left[-\\log_{}\\left|t\\right|+\\log_{}\\left|t-1\\right|\\right]+C}\\\\\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601293440330,"cs":"6zMZddQ4vgow5vn6UDHtpg==","size":{"width":282,"height":36}}$

Put the value of t:-

${"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"14-0-1-1-0-1-0-1-1-3-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{n}\\left[-\\log_{}\\left|x^{n}+1\\right|+\\log_{}\\left|x^{n}+1-1\\right|\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{n}\\left[-\\log_{}\\left|x^{n}+1\\right|+\\log_{}\\left|x^{n}\\right|\\right]+C}\t\n\\end{align*}","ts":1601293427692,"cs":"R4Bgi2t63N91zWxsvZoLdQ==","size":{"width":358,"height":76}}$

We know that:-

log m - log n = log (m/n)

${"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{n}\\log_{}\\left|\\frac{x^{n}}{x^{n}+1}\\right|+C}\t\n\\end{align*}","type":"align*","id":"14-0-1-1-0-1-0-1-1-3-1-0","ts":1601293407071,"cs":"ApiplpE8DpRmtTqadt3lgA==","size":{"width":220,"height":36}}$

${"code":"$17.\\,\\frac{\\cos x}{\\left(1-\\sin x\\right)\\left(2-\\sin x\\right)}$","type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-2-1-0","ts":1601293535285,"cs":"doLE9ggSGCpxo5JnPBKdNg==","size":{"width":153,"height":25}}$

Solun:- Let f(x) = ${"code":"$\\frac{\\cos x}{\\left(1-\\sin x\\right)\\left(2-\\sin x\\right)}$","id":"5-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-0","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601293549753,"cs":"HoVNv/iz3f5j/0Kxv6RLqA==","size":{"width":96,"height":20}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\cos x}{\\left(1-\\sin x\\right)\\left(2-\\sin x\\right)}.dx}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-1-0","ts":1601293587192,"cs":"bGm+LiqlOLZfNKvK8FQKuw==","size":{"width":280,"height":36}}$

Let sin x = t

Differentiate w.r.t. to t:-

⇒ cos x.dx = dt

${"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-1-1","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{\\left(1-t\\right)\\left(2-t\\right)}}\t\n\\end{align*}","ts":1601293774029,"cs":"ApfXTdm0lEYi5N90Cbr4bg==","size":{"width":204,"height":36}}$

By partial Fraction:-

${"font":{"color":"#000000","size":12,"family":"Arial"},"type":"$","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-0","code":"$\\frac{1}{\\left(1-t\\right)\\left(2-t\\right)}=\\frac{A}{1-t}+\\frac{B}{2-t}$","ts":1601293856966,"cs":"f8IaS0LYZ7BZXuke6Ovo5A==","size":{"width":196,"height":28}}$

⇒ 1 = A(2 - t) + B(1 - t)

Put t = 1:-

⇒ 1 = A(1)

⇒ A = 1

Put t = 2:-

⇒ 1 = B(-1)

⇒ B = -1 then

${"type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"code":"$\\frac{1}{\\left(1-t\\right)\\left(2-t\\right)}=\\frac{1}{1-t}-\\frac{1}{2-t}$","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-1","ts":1601294123453,"cs":"j/XCqqCtKQ+luwXQQrJTsg==","size":{"width":196,"height":28}}$

${"id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-1-1-2-1-0","font":{"size":8,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{1}{1-t}-\\frac{1}{2-t}\\right].dt}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{1-t}.dt-\\int_{}^{}\\frac{1}{2-t}.dt}\t\n\\end{align*}","type":"align*","ts":1601553480917,"cs":"6Pm+tNvaZFeZS9Je4gIJHw==","size":{"width":229,"height":66}}$

We know that:-

${"id":"3-0-0-0-0-0-5-1-3-1-4","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601110014037,"cs":"03lCKNwgxVRNwL45FD8LhQ==","size":{"width":140,"height":18}}$

${"id":"14-0-1-1-0-1-0-1-1-2-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\log_{}\\left|1-t\\right|-\\left(-\\log_{}\\left|2-t\\right|\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|2-t\\right|-\\log_{}\\left|1-t\\right|+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601294433441,"cs":"fu4iRpzlqr4KnC7mYL+XVg==","size":{"width":312,"height":76}}$

Put the value of t:-

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"14-0-1-1-0-1-0-1-1-2-1-1","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|2-\\sin x\\right|-\\log_{}\\left|1-\\sin x\\right|+C}\t\n\\end{align*}","ts":1601294467509,"cs":"0LHQPzPGgEr4x22dgQfPNw==","size":{"width":320,"height":36}}$

We know that:-

log m - log n = log (m/n)

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\frac{2-\\sin x}{1-\\sin x}\\right|+C}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"14-0-1-1-0-1-0-1-1-3-1-1","type":"align*","ts":1601294498922,"cs":"kySJNndeHNsmdCf3TSdisA==","size":{"width":218,"height":36}}$

${"font":{"color":"#000000","family":"Arial","size":12},"type":"$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-2-1-1","code":"$18.\\,\\frac{\\left(x^{2}+1\\right)\\left(x^{2}+2\\right)}{\\left(x^{2}+3\\right)\\left(x^{2}+4\\right)}$","ts":1601376361079,"cs":"Z4A00HusuOxXpIYG6adsrg==","size":{"width":129,"height":34}}$

Solun:- Let f(x) = ${"id":"5-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","code":"$\\frac{\\left(x^{2}+1\\right)\\left(x^{2}+2\\right)}{\\left(x^{2}+3\\right)\\left(x^{2}+4\\right)}$","font":{"color":"#000000","family":"Arial","size":10},"ts":1601376418591,"cs":"XwVGmz8UzX71o9fI7HM/YQ==","size":{"width":77,"height":28}}$

Integrate f(x):-

${"font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-1-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\left(x^{2}+1\\right)\\left(x^{2}+2\\right)}{\\left(x^{2}+3\\right)\\left(x^{2}+4\\right)}.dx}\t\n\\end{align*}","type":"align*","ts":1601376481390,"cs":"N4eUtUTBk8/hOPtHXDZOIA==","size":{"width":252,"height":41}}$

Put x2 = t

${"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\left(t+1\\right)\\left(t+2\\right)}{\\left(t+3\\right)\\left(t+4\\right)}.dt}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{t^{2}+3t+2}{t^{2}+7t+12}.dt}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601381119960,"cs":"zeWJ0bGj+4GhUrcap9XTtA==","size":{"width":225,"height":80}}$

${"id":"16","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\ldiv{t^{2}+7t+12}{t^{2}+3t+2}$","type":"$","ts":1601376745924,"cs":"+RfjLtaz9T7GpVg01LcAsQ==","size":{"width":165,"height":24}}$

By quotient rule:-

${"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"8-0-1","code":"\\begin{align*}\n{t^{2}+3t+2}&={\\left(t^{2}+7t+12\\right).\\left(1\\right)+\\left(-4t-10\\right)}\\\\\n{t^{2}+3t+2}&={\\left(t^{2}+7t+12\\right).\\left(1\\right)-\\left(4t+10\\right)}\t\n\\end{align*}","ts":1601377011111,"cs":"gg53GCpzgvt0AI5FoMyrwQ==","size":{"width":308,"height":44}}$

${"font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\frac{t^{2}+3t+2}{t^{2}+7t+12}}&={\\frac{t^{2}+7t+12}{t^{2}+7t+12}-\\frac{4t+10}{t^{2}+7t+12}}\\\\\n{\\frac{t^{2}+3t+2}{t^{2}+7t+12}}&={1-\\frac{4t+10}{t^{2}+7t+12}}\t\n\\end{align*}","id":"9-0-1","type":"align*","ts":1601377184295,"cs":"L8A/LloKxrWIOc6bNG2Fhg==","size":{"width":301,"height":77}}$

${"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(1-\\frac{4t+10}{t^{2}+7t+12}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}1.dx-2\\int_{}^{}\\frac{2t+5}{t^{2}+7t+12}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}1.dx-2\\int_{}^{}\\frac{2t+5}{\\left(t+3\\right)\\left(t+4\\right)}.dx}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601550115216,"cs":"tnxzjpCJ9527TU5EFmYl9w==","size":{"width":308,"height":120}}$

By partial Fraction:-

${"type":"$","code":"$\\frac{2t+5}{\\left(t+3\\right)\\left(t+4\\right)}=\\frac{A}{t+3}+\\frac{B}{t+4}$","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-0-0","font":{"family":"Arial","size":12,"color":"#000000"},"ts":1601550232284,"cs":"D06qpTMYEV/LS4eVconYxA==","size":{"width":196,"height":28}}$

⇒ 2t + 5 = A(t + 4) + B(t + 3)

Put t = - 4:-

⇒ - 3 = B(- 1)

⇒ B = 3

Put t = - 3:-

⇒ - 1 = A(1)

⇒ A = -1 then

${"code":"$\\frac{2t+5}{\\left(t+3\\right)\\left(t+4\\right)}=\\frac{-1}{t+3}+\\frac{3}{t+4}$","type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-1","ts":1601550292685,"cs":"QWpRf9+ETMaTFPhlPHlp0A==","size":{"width":196,"height":28}}$

${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}1.dx-2\\int_{}^{}\\left(\\frac{-1}{t+3}+\\frac{3}{t+4}\\right).dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1","ts":1601550323077,"cs":"mGJ22ELjm4ulv1DMycUcHA==","size":{"width":330,"height":37}}$

Put t = x2

${"type":"align*","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-1-1-2-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}1.dx-2\\int_{}^{}\\frac{-1}{x^{2}+3}+\\frac{3}{x^{2}+4}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}1.dx+\\int_{}^{}\\frac{2}{x^{2}+3}.dx-\\int_{}^{}\\frac{6}{x^{2}+4}.dx}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601378591733,"cs":"dYsbWDhWUwcnfmuuVKoL2Q==","size":{"width":362,"height":76}}$

We know that:-

${"type":"$","code":"$\\int_{}^{}\\frac{1}{x^{2}+a^{2}}.dx=\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+c$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"3-0-0-0-0-0-5-1-3-1-5-0","ts":1601378704941,"cs":"yIsDHeQNb49F7R0oAdo+nw==","size":{"width":204,"height":20}}$

${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x+\\frac{2}{{\\sqrt[]{3}}}\\tan^{-1}\\left(\\frac{x}{{\\sqrt[]{3}}}\\right)-\\frac{6}{2}\\tan^{-1}\\left(\\frac{x}{2}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x+\\frac{2}{{\\sqrt[]{3}}}\\tan^{-1}\\left(\\frac{x}{{\\sqrt[]{3}}}\\right)-3\\tan^{-1}\\left(\\frac{x}{2}\\right)+C}\t\n\\end{align*}","id":"14-0-1-1-0-1-0-1-1-2-1-2-0","type":"align*","ts":1601378857322,"cs":"MgoI5x7SCdHRcPxWqVJDWg==","size":{"width":377,"height":82}}$

${"code":"$19.\\,\\frac{2x}{\\left(x^{2}+1\\right)\\left(x^{2}+3\\right)}$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-2-1-2-0","ts":1601380886873,"cs":"XuKddRgSR6NIIA0RuXp+VQ==","size":{"width":125,"height":28}}$

Solun:- Let f(x) = ${"code":"$\\frac{2x}{\\left(x^{2}+1\\right)\\left(x^{2}+3\\right)}$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"5-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","ts":1601380957994,"cs":"2NbC+obdj9owxled2+pb3A==","size":{"width":74,"height":21}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{2x}{\\left(x^{2}+1\\right)\\left(x^{2}+3\\right)}.dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-1-1-0-1-0","type":"align*","ts":1601380991894,"cs":"aIiHSNmUN4tBHeC5J2C6GQ==","size":{"width":248,"height":36}}$

Let x2 = t

Differentiate w.r.t. to t:-

2x.dx = dt

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{\\left(t+1\\right)\\left(t+3\\right)}}\\\\\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1601381082995,"cs":"gEyrSIxQ0eE88SI7WUvCOQ==","size":{"width":204,"height":36}}$

By partial Fraction:-

${"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-0-1-0","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"code":"$\\frac{1}{\\left(t+1\\right)\\left(t+3\\right)}=\\frac{A}{t+1}+\\frac{B}{t+3}$","ts":1601381243056,"cs":"Nz4pI11J0tun2RhXPx4Fcg==","size":{"width":196,"height":28}}$

⇒ 1 = A(t + 3) + B(t + 1)

Put t = - 1:-

⇒ 1 = A(2)

⇒ A = 1/2

Put t = - 3:-

⇒ 1 = B(-2)

⇒ B = -1/2 then

${"type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-0-1-1","code":"$\\frac{1}{\\left(t+1\\right)\\left(t+3\\right)}=\\frac{1}{2}\\frac{1}{t+1}-\\frac{1}{2}\\frac{1}{t+3}$","ts":1601381469398,"cs":"Wop5B897FaEJiXUu9ZzGrg==","size":{"width":228,"height":28}}$

${"id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-1-1-2-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{1}{2}\\frac{1}{t+1}-\\frac{1}{2}\\frac{1}{t+3}\\right).dt}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\frac{1}{t+1}.dt-\\frac{1}{2}\\int_{}^{}\\frac{1}{t+3}.dt}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","ts":1601553530898,"cs":"1/r/OUqzfEns+sxtXfb7wA==","size":{"width":302,"height":77}}$

We know that:-

${"font":{"color":"#000000","size":10,"family":"Arial"},"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","id":"3-0-0-0-0-0-5-1-3-1-5-1","ts":1601381618045,"cs":"k8MI5q6mYnnExTcbBD82Gw==","size":{"width":140,"height":18}}$

${"id":"14-0-1-1-0-1-0-1-1-2-1-2-1-0-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left|t+1\\right|-\\frac{1}{2}\\log_{}\\left|t+3\\right|+C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601381774227,"cs":"g79T8CctDsetTtPDnKZDjQ==","size":{"width":298,"height":36}}$

Put the value of t:-

${"type":"align*","id":"14-0-1-1-0-1-0-1-1-2-1-2-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left|x^{2}+1\\right|-\\frac{1}{2}\\log_{}\\left|x^{2}+3\\right|+C}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1601381831158,"cs":"juVN6HifTGAQUXd0hzcfCg==","size":{"width":321,"height":36}}$

We know that:-

log m - log n = log (m/n)

${"id":"17-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left|\\frac{x^{2}+1}{x^{2}+3}\\right|+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601381922045,"cs":"RuVmOAH1Ei5d4LG6VqenZg==","size":{"width":217,"height":40}}$

${"font":{"color":"#000000","family":"Arial","size":12},"code":"$20.\\,\\frac{1}{x\\left(x^{4}-1\\right)}$","type":"$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-2-1-2-1","ts":1601382129048,"cs":"R2OMLBN8EeH8bk1UV51iRw==","size":{"width":89,"height":28}}$

Solun:- Let f(x) = ${"code":"$\\frac{1}{x\\left(x^{4}-1\\right)}$","id":"5-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601382152896,"cs":"+AxP9+9680PFkqhvmEytWg==","size":{"width":46,"height":22}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{x\\left(x^{4}-1\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{3}}{x^{3}.x\\left(x^{4}-1\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{3}}{x^{4}\\left(x^{4}-1\\right)}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-1-1-0-1-1-0","ts":1601382404339,"cs":"mVLfJB/z6101Nlqom9lJig==","size":{"width":224,"height":124}}$

Let x4 = t

Differentiate w.r.t. to t:-

⇒ 4x3.dx = dt

⇒ x3.dx = (1/4).dt

${"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-1-1-1-1-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{t\\left(t-1\\right)}.\\frac{dt}{4}}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\int_{}^{}\\frac{1}{t\\left(t-1\\right)}.dt}\t\n\\end{align*}","ts":1601383288132,"cs":"cTjo4Z1Bpxgt99JSa0sXeA==","size":{"width":204,"height":77}}$

By partial Fraction:-

${"code":"$\\frac{1}{t\\left(t-1\\right)}=\\frac{A}{t}+\\frac{B}{t-1}$","type":"$","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-0-1-2-0","font":{"family":"Arial","color":"#000000","size":12},"ts":1601383636823,"cs":"CjgqozaeKLgw671hGnatVw==","size":{"width":152,"height":28}}$

⇒ 1 = A(t - 1) + B(t)

Put t = 1:-

⇒ 1 = B(1)

⇒ B = 1

Put t = 0:-

⇒ 1 = A(-1)

⇒ A = -1 then

${"font":{"color":"#000000","family":"Arial","size":12},"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-0-1-2-1","type":"$","code":"$\\frac{1}{t\\left(t-1\\right)}=\\frac{-1}{t}+\\frac{1}{t-1}$","ts":1601551127605,"cs":"Eb8hC2TGk94HB73JPSJevw==","size":{"width":160,"height":28}}$

${"font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-1-1-2-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\int_{}^{}\\left(\\frac{-1}{t}+\\frac{1}{t-1}\\right).dt}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\left[\\int_{}^{}\\frac{1}{t-1}.dt-\\int_{}^{}\\frac{1}{t}.dt\\right]}\t\n\\end{align*}","ts":1601553565960,"cs":"4qpdMovRHUE5+rMHkiKIGw==","size":{"width":274,"height":80}}$

We know that:-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\left[\\log_{}\\left|t-1\\right|-\\log_{}\\left|t\\right|\\right]+C}\t\n\\end{align*}","id":"14-0-1-1-0-1-0-1-1-2-1-2-1-0-1","font":{"family":"Arial","color":"#000000","size":10},"ts":1601551419622,"cs":"Bi8+SejddV9nN7gEZ1Ymzg==","size":{"width":265,"height":36}}$

Put the value of t:-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\left[\\log_{}\\left|x^{4}-1\\right|-\\log_{}\\left|x^{4}\\right|\\right]+C}\t\n\\end{align*}","id":"14-0-1-1-0-1-0-1-1-2-1-2-1-1-1","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601551441102,"cs":"WE2L3bjuRXhweP5HMZpIcw==","size":{"width":292,"height":36}}$

We know that:-

log m - log n = log (m/n)

${"font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\log_{}\\left|\\frac{x^{4}-1}{x^{4}}\\right|+C}\t\n\\end{align*}","type":"align*","id":"17-1","ts":1601551464521,"cs":"HBmWcuj1olJpUtcYKxQhCA==","size":{"width":217,"height":40}}$

${"code":"$21.\\,\\frac{1}{\\left(e^{x}-1\\right)}$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-2-1-2-1","ts":1601454407224,"cs":"SUjOjvj7FoDDov3iV4/hAA==","size":{"width":80,"height":28}}$

Solun:- Let f(x) = ${"id":"5-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$\\frac{1}{\\left(e^{x}-1\\right)}$","ts":1601454442200,"cs":"yNG3EdaXpAkVJZy2y8NLpw==","size":{"width":40,"height":21}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-1-1-0-1-1-1-0","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\left(e^{x}-1\\right)}.dx}\t\n\\end{align*}","ts":1601454552469,"cs":"T0woXO/jzt/In/kh35yP5A==","size":{"width":192,"height":36}}$

Let ex = t

Differentiate w.r.t. to t:-

⇒ ex.dx = dt

⇒ dx = (1/ex).dt

⇒ dx = (1/t).dt

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\left(t-1\\right)}.\\frac{dt}{t}}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{t\\left(t-1\\right)}.dt}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-1-1-0-1-1-1-1","ts":1601454735257,"cs":"iWF1u5vNFeIDwNiYPcX0xQ==","size":{"width":188,"height":77}}$

By partial Fraction:-

${"code":"$\\frac{1}{t\\left(t-1\\right)}=\\frac{A}{t}+\\frac{B}{t-1}$","type":"$","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-0-1-2-2-0","font":{"family":"Arial","color":"#000000","size":12},"ts":1601383636823,"cs":"//jCuGygjabpjuolGu7D8w==","size":{"width":152,"height":28}}$

⇒ 1 = A(t - 1) + B(t)

Put t = 1:-

⇒ 1 = B(1)

⇒ B = 1

Put t = 0:-

⇒ 1 = A(-1)

⇒ A = -1 then

${"type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"code":"$\\frac{1}{t\\left(t-1\\right)}=\\frac{-1}{t}+\\frac{1}{t-1}$","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-0-1-2-3","ts":1601455229934,"cs":"rKeH/R6SlmxkD/07IDVHtA==","size":{"width":160,"height":28}}$

${"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{-1}{t}+\\frac{1}{t-1}\\right).dt}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{t-1}.dt-\\int_{}^{}\\frac{1}{t}.dt}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-1-1-2-1-1-1-1-1-0","ts":1601553630303,"cs":"fnBXKNi0Z95+4h3OBfrsUw==","size":{"width":242,"height":77}}$

We know that:-

${"font":{"family":"Arial","color":"#000000","size":10},"id":"14-0-1-1-0-1-0-1-1-2-1-2-1-0-2-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|t-1\\right|-\\log_{}\\left|t\\right|+C}\t\n\\end{align*}","ts":1601455310066,"cs":"31oDqnl4y6qNyQj/lPyXoQ==","size":{"width":241,"height":36}}$

Put the value of t:-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|e^{x}-1\\right|-\\log_{}\\left|e^{x}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\frac{e^{x}-1}{e^{x}}\\right|+C}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"14-0-1-1-0-1-0-1-1-2-1-2-1-1-2","ts":1601455371972,"cs":"4TI7LgxxCOQKmozuhXpLcA==","size":{"width":260,"height":76}}$

Choose the correct answer in each of the Exercises 22 and 23.

${"font":{"color":"#000000","size":12,"family":"Arial"},"type":"$","code":"$22.\\,\\int_{}^{}\\frac{x.dx}{\\left(x-1\\right)\\left(x-2\\right)}$","id":"1-0-0-0-0-1-0","ts":1601456326783,"cs":"ZjVRd2RlW+dYs2lZaMIvJQ==","size":{"width":134,"height":28}}$

Solun:- Let f(x) = ${"font":{"family":"Arial","size":10,"color":"#000000"},"id":"5-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","code":"$\\frac{x}{\\left(x-1\\right)\\left(x-2\\right)}$","type":"$","ts":1601456558338,"cs":"EeYwPhAacDDdk1EAZ+CykQ==","size":{"width":65,"height":20}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x.dx}{\\left(x-1\\right)\\left(x-2\\right)}}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-1-1-0-1-1-1-2-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1601456531115,"cs":"NOstTddZrfjezWzc3n9Z4Q==","size":{"width":212,"height":36}}$

By partial Fraction:-

${"font":{"family":"Arial","size":12,"color":"#000000"},"code":"$\\frac{x}{\\left(x-1\\right)\\left(x-2\\right)}=\\frac{A}{\\left(x-1\\right)}+\\frac{B}{\\left(x-2\\right)}$","id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-0-1-2-2-1-0-0","type":"$","ts":1601458462187,"cs":"w+PP2RBy4wumqdcvzj+8rQ==","size":{"width":229,"height":28}}$

⇒ x = A(x - 2) + B(x - 1)

Put x = 1:-

⇒ 1 = A(-1)

⇒ A = - 1

Put x = 2:-

⇒ 2 = B(1)

⇒ B = 2 then

${"code":"$\\frac{x}{\\left(x-1\\right)\\left(x-2\\right)}=\\frac{-1}{\\left(x-1\\right)}+\\frac{2}{\\left(x-2\\right)}$","font":{"family":"Arial","size":12,"color":"#000000"},"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-0-1-2-2-1-1-0","type":"$","ts":1601458427716,"cs":"RTC+y8kYy06UG8JQdGhzrA==","size":{"width":229,"height":28}}$

${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{-1}{\\left(x-1\\right)}+\\frac{2}{\\left(x-2\\right)}\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={2\\int_{}^{}\\frac{1}{\\left(x-2\\right)}.dx-\\int_{}^{}\\frac{1}{\\left(x-1\\right)}.dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-1-1-2-1-1-1-1-1-1-0","ts":1601553667768,"cs":"M6f+d/IcRHQbZPFT8z613w==","size":{"width":316,"height":80}}$

We know that:-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={2\\log_{}\\left|x-2\\right|-\\log_{}\\left|x-1\\right|+C}\t\n\\end{align*}","id":"14-0-1-1-0-1-0-1-1-2-1-2-1-0-2-1-0-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1601457665186,"cs":"K7pAUWBi90OlpZwwOLN5zg==","size":{"width":286,"height":36}}$

We know that:- n.log m = log mn

log m - log n = log (m/n)

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\left(x-2\\right)^{2}\\right|-\\log_{}\\left|x-1\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\frac{\\left(x-2\\right)^{2}}{x-1}\\right|+C}\t\n\\end{align*}","type":"align*","font":{"size":8.678571428571429,"family":"Arial","color":"#000000"},"id":"14-0-1-1-0-1-0-1-1-2-1-2-1-0-2-1-1-0","ts":1601457944005,"cs":"eXMPOzRj03ZolzkvUcFW/w==","size":{"width":296,"height":81}}$

The correct answer is B.

${"code":"$23.\\,\\int_{}^{}\\frac{dx}{x\\left(x^{2}+1\\right)}$","font":{"color":"#000000","family":"Arial","size":12},"id":"1-0-0-0-0-1-1","type":"$","ts":1601458210585,"cs":"X93gYRZBpzvR/HosMyhm5w==","size":{"width":112,"height":28}}$

Solun:- Let f(x) = ${"code":"$\\frac{1}{x\\left(x^{2}+1\\right)}$","id":"5-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","ts":1601458380580,"cs":"/J1dsKv4Bk1m39WHchSEXg==","size":{"width":46,"height":21}}$

Integrate f(x):-

${"font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-1-1-1-1-1-0-0-1-1-1-1-1-1-1-1-1-0-0-1-1-0-1-1-1-2-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dx}{x\\left(x^{2}+1\\right)}}\t\n\\end{align*}","type":"align*","ts":1601458340863,"cs":"AQTmhUzQs7K1Bng5dBY76g==","size":{"width":178,"height":36}}$

By partial Fraction:-

${"type":"$","font":{"family":"Arial","color":"#000000","size":12},"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-0-1-2-2-1-0-1-0","code":"$\\frac{1}{x\\left(x^{2}+1\\right)}=\\frac{A}{x}+\\frac{Bx+C}{\\left(x^{2}+1\\right)}$","ts":1601458603142,"cs":"semr6+QQUeuGsj6EP/iQTg==","size":{"width":184,"height":29}}$

⇒ 1 = A(x2 + 1) + (Bx+C)(x)

Put x = 0:-

⇒ 1 = A(1)

⇒ A = 1

Put x = 1:-

⇒ 1 = 1(2) + (B+C)(1)

⇒ B + C = - 1

Put x = - 1:-

⇒ 1 = 1(2) - (- B + C)(1)

⇒ - B + C = 1

⇒ C = 0 and B = - 1 then

${"font":{"family":"Arial","size":12,"color":"#000000"},"id":"6-0-1-0-1-1-0-1-0-0-0-1-0-1-0-1-0-1-0-1-0-1-0-1-0-2-0-1-0-1-0-0-1-2-2-1-0-1-1","type":"$","code":"$\\frac{1}{x\\left(x^{2}+1\\right)}=\\frac{1}{x}-\\frac{x}{\\left(x^{2}+1\\right)}$","ts":1601460210972,"cs":"NL3fA9yhvDxfeDnvsEQZQA==","size":{"width":182,"height":28}}$

${"id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-1-1-2-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{1}{x}-\\frac{x}{\\left(x^{2}+1\\right)}\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{x}.dx-\\int_{}^{}\\frac{x}{\\left(x^{2}+1\\right)}.dx}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1601553796736,"cs":"G/MAwSUJkbOSH+/0W1d8pA==","size":{"width":272,"height":80}}$

Let x2 + 1 = t

Differentiate w.r.t. to t:-

⇒ 2x.dx = dt

⇒ x.dx = (1/2).dt

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{x}.dx-\\int_{}^{}\\frac{1}{t}.\\frac{dt}{2}}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{x}.dx-\\frac{1}{2}\\int_{}^{}\\frac{1}{t}.dt}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"2-0-0-0-0-0-0-0-1-1-1-1-1-1-0-1-1-1-1-1-0-1-1-2-1-1-1-1-1-1-1-1","ts":1601461307469,"cs":"Wiom6Um0zQf8nz0Y5HLCwA==","size":{"width":240,"height":76}}$

We know that:-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|x\\right|-\\frac{1}{2}\\log_{}\\left|t\\right|+C}\t\n\\end{align*}","id":"14-0-1-1-0-1-0-1-1-2-1-2-1-0-2-1-0-1","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601461694784,"cs":"4YpdVGRWz880zZFygi0CCA==","size":{"width":232,"height":36}}$

${"id":"14-0-1-1-0-1-0-1-1-2-1-2-1-0-2-1-1-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|x\\right|-\\frac{1}{2}\\log_{}\\left|x^{2}+1\\right|+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1601461975076,"cs":"KZa1LOBrJR2yHcxMMBuZjw==","size":{"width":272,"height":36}}$

The correct answer is A.

Download PDF of Exercise 7.5 Solutions

NCERT Maths solutions for 12th

Important Note

Exercise 7.5

Integrate the functions in Exercises 1 to 21.

${"type":"$","id":"1-0-0-0-0-0","code":"$1.\\,\\frac{x}{\\left(x+1\\right)\\left(x+2\\right)}$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1601109030876,"cs":"S8tA0V7J/w9i/kZ+zC45pg==","size":{"width":104,"height":25}}$

${"font":{"family":"Arial","size":12,"color":"#000000"},"code":"$2.\\,\\frac{1}{x^{2}-9}$","id":"1-0-0-0-1-0","type":"$","ts":1601111318685,"cs":"p/OBnu7jDDywErxZZWS/LA==","size":{"width":60,"height":25}}$

${"type":"$","id":"1-0-0-0-1-1-0","font":{"color":"#000000","family":"Arial","size":12},"code":"$3.\\,\\frac{3x-1}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}$","ts":1601112384262,"cs":"oXPcFQoXrFo8BTdvZvnQrw==","size":{"width":141,"height":28}}$

${"type":"$","code":"$4.\\,\\frac{x}{\\left(x-1\\right)\\left(x-2\\right)\\left(x-3\\right)}$","id":"1-0-0-0-1-1-1-0","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1601114431410,"cs":"Wk7kMdA1gGDcRMFsPkeFhg==","size":{"width":141,"height":25}}$

${"code":"$5.\\,\\frac{2x}{x^{2}+3x+2}$","id":"1-0-0-0-1-1-1-1-0","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1601116447020,"cs":"iKy+kJO3hXNmRzY6RFf0ig==","size":{"width":86,"height":25}}$

${"id":"1-0-0-0-1-1-1-1-1-0-0","type":"$","font":{"family":"Arial","color":"#000000","size":12},"code":"$6.\\,\\frac{1-x^{2}}{x\\left(1-2x\\right)}$","ts":1601117865801,"cs":"y0ECPGo1ucxhL/sXkmblcw==","size":{"width":80,"height":30}}$

${"code":"$7.\\,\\frac{x}{\\left(x^{2}+1\\right)\\left(x-1\\right)}$","type":"$","id":"1-0-0-0-1-1-1-1-1-1","font":{"color":"#000000","family":"Arial","size":12},"ts":1601121088377,"cs":"QDQ3MSqabQ+1sRolu9GVQg==","size":{"width":109,"height":25}}$

${"font":{"family":"Arial","size":12,"color":"#000000"},"type":"$","code":"$8.\\,\\frac{x}{\\left(x-1\\right)^{2}\\left(x+2\\right)}$","id":"1-0-0-0-1-1-1-1-1-0-1-0","ts":1601197718329,"cs":"YXpwG5xbrF/lrGePp2jdHQ==","size":{"width":109,"height":28}}$

${"id":"1-0-0-0-1-1-1-1-1-0-1-1-0","type":"$","code":"$9.\\,\\frac{3x+5}{x^{3}-x^{2}-x+1}$","font":{"color":"#000000","size":12,"family":"Arial"},"ts":1601200007652,"cs":"7jDrgIfx50UT4JPilWzfew==","size":{"width":104,"height":26}}$

${"type":"$","code":"$10.\\,\\frac{2x-3}{\\left(x^{2}-1\\right)\\left(2x+3\\right)}$","font":{"color":"#000000","family":"Arial","size":12},"id":"1-0-0-0-1-1-1-1-1-0-1-1-1-0","ts":1601201291672,"cs":"ATOAEh0F2YP4HMnMFQF00g==","size":{"width":126,"height":28}}$

${"code":"$11.\\,\\frac{5x}{\\left(x+1\\right)\\left(x^{2}-4\\right)}$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-0","type":"$","font":{"family":"Arial","color":"#000000","size":12},"ts":1601202828333,"cs":"xnbsMmFbSd/Vhm9XRW5UFg==","size":{"width":120,"height":28}}$

${"code":"$12.\\,\\frac{x^{3}+x+1}{x^{2}-1}$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":12},"type":"$","ts":1601203979768,"cs":"UPQui5AYGCaS24ktZXJEPQ==","size":{"width":89,"height":28}}$

${"code":"$13.\\,\\frac{2}{\\left(1-x\\right)\\left(1+x^{2}\\right)}$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1601205904041,"cs":"kJ/u1s8MMeArGI6moBo8kw==","size":{"width":120,"height":28}}$

${"type":"$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":12},"code":"$14.\\,\\frac{3x-1}{\\left(x+2\\right)^{2}}$","ts":1601209038963,"cs":"gfCqcaLo7lQYPJ/N7Kt04A==","size":{"width":81,"height":32}}$

${"font":{"family":"Arial","size":12,"color":"#000000"},"code":"$15.\\,\\frac{1}{x^{4}-1}$","type":"$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1","ts":1601286770720,"cs":"8IZeqLREUGGXVBwNq9u8rw==","size":{"width":70,"height":26}}$

${"type":"$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-2-0","code":"$16.\\,\\frac{1}{x\\left(x^{n}+1\\right)}$","font":{"color":"#000000","size":12,"family":"Arial"},"ts":1601291984018,"cs":"1oLUxwOBUzh+ii2L0TsbCw==","size":{"width":90,"height":28}}$

${"code":"$17.\\,\\frac{\\cos x}{\\left(1-\\sin x\\right)\\left(2-\\sin x\\right)}$","type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-2-1-0","ts":1601293535285,"cs":"doLE9ggSGCpxo5JnPBKdNg==","size":{"width":153,"height":25}}$

${"code":"$19.\\,\\frac{2x}{\\left(x^{2}+1\\right)\\left(x^{2}+3\\right)}$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-2-1-2-0","ts":1601380886873,"cs":"XuKddRgSR6NIIA0RuXp+VQ==","size":{"width":125,"height":28}}$

${"font":{"color":"#000000","family":"Arial","size":12},"code":"$20.\\,\\frac{1}{x\\left(x^{4}-1\\right)}$","type":"$","id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-2-1-2-1","ts":1601382129048,"cs":"R2OMLBN8EeH8bk1UV51iRw==","size":{"width":89,"height":28}}$

${"code":"$21.\\,\\frac{1}{\\left(e^{x}-1\\right)}$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"1-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-2-1-2-1","ts":1601454407224,"cs":"SUjOjvj7FoDDov3iV4/hAA==","size":{"width":80,"height":28}}$

Choose the correct answer in each of the Exercises 22 and 23.

${"font":{"color":"#000000","size":12,"family":"Arial"},"type":"$","code":"$22.\\,\\int_{}^{}\\frac{x.dx}{\\left(x-1\\right)\\left(x-2\\right)}$","id":"1-0-0-0-0-1-0","ts":1601456326783,"cs":"ZjVRd2RlW+dYs2lZaMIvJQ==","size":{"width":134,"height":28}}$

${"code":"$23.\\,\\int_{}^{}\\frac{dx}{x\\left(x^{2}+1\\right)}$","font":{"color":"#000000","family":"Arial","size":12},"id":"1-0-0-0-0-1-1","type":"$","ts":1601458210585,"cs":"X93gYRZBpzvR/HosMyhm5w==","size":{"width":112,"height":28}}$

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Important Note

Integrals(Ex. 7.5)

Exercise 7.5

Integrate the functions in Exercises 1 to 21.

Choose the correct answer in each of the Exercises 22 and 23.

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