Exercise 7.4

Integrate the functions in Exercises 1 to 23.

${"id":"1-0-0-0","type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"code":"$1.\\,\\frac{3x^{2}}{x^{6}+1}$","ts":1600679911258,"cs":"yhhcytWV2Zy+8aue22/oIg==","size":{"width":60,"height":28}}$

Solun:- Let f(x) = ${"type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$\\frac{3x^{2}}{x^{6}+1}$","id":"5-0","ts":1600679950266,"cs":"353HFo4EPs8mAGje1eRwZw==","size":{"width":32,"height":21}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-0-0","type":"$","font":{"family":"Arial","color":"#000000","size":12},"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\frac{3x^{2}}{x^{6}+1}.dx$","ts":1600680107429,"cs":"SeibGKQB9W99A9VrjJpsDw==","size":{"width":196,"height":28}}$

Let x3 = t

Differentiate w.r.t to t:-

⇒ 3x2.dx = dt

${"id":"2-1-0-0-0-0-0","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{t^{2}+1}}\t\n\\end{align*}","ts":1600680787922,"cs":"acTJvcyE/IZJCnhdrpovgw==","size":{"width":153,"height":36}}$

We know that:-

${"type":"$","font":{"color":"#000000","family":"Arial","size":10},"code":"$\\int_{}^{}\\frac{1}{1+x^{2}}.dx=\\tan^{-1}x+c$","id":"3-0-0-0-0-0","ts":1600680867740,"cs":"xfwZe1rdov2AS2y7ps493g==","size":{"width":168,"height":20}}$

${"id":"2-1-1-0-0-0-0-0-0-0","font":{"family":"Arial","color":"#000000","size":10},"type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\tan^{-1}t+C$","ts":1600680852552,"cs":"r0sgCPeyGQ8R+CB4G9qaQw==","size":{"width":160,"height":18}}$

Put the value of t:-

${"id":"2-1-1-0-0-1-0-0-0","type":"$","font":{"color":"#000000","family":"Arial","size":10},"code":"$\\int_{}^{}f\\left(x\\right)dx=\\tan^{-1}\\left(x^{3}\\right)+C$","ts":1600681026079,"cs":"B0AdfQHhFsOLyiedcu3f2g==","size":{"width":184,"height":18}}$

${"font":{"size":12,"family":"Arial","color":"#000000"},"code":"$2.\\,\\frac{1}{{\\sqrt[]{1+4x^{2}}}}$","id":"1-1-0","type":"$","ts":1600681152705,"cs":"yH+4L+p5Pz3pKEeeX+74fg==","size":{"width":80,"height":29}}$

Solun:- Let f(x) = ${"font":{"family":"Arial","color":"#000000","size":10},"code":"$\\frac{1}{{\\sqrt[]{1+4x^{2}}}}$","type":"$","id":"6-0","ts":1600681222191,"cs":"WfSuHM2p6O+Z7LKbwBw1Zw==","size":{"width":46,"height":22}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{1+4x^{2}}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{4\\left(\\frac{1}{4}+x^{2}\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(\\frac{1}{4}+x^{2}\\right)}}}.dx}\t\n\\end{align*}","type":"align*","ts":1600681399947,"cs":"ftoi94OlCQrkorBQOrigGg==","size":{"width":233,"height":148}}$

We know that:-

${"type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}+a^{2}}}\\right|+c$","id":"3-0-0-0-1-0-0","ts":1600681532828,"cs":"i3LZV6DZlhoxOTicUEO3Wg==","size":{"width":257,"height":24}}$

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left|x+{\\sqrt[]{x^{2}+\\frac{1}{4}}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left|x+{\\sqrt[]{\\frac{4x^{2}+1}{4}}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left|x+\\frac{1}{2}{\\sqrt[]{4x^{2}+1}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left|\\frac{2x+{\\sqrt[]{4x^{2}+1}}}{2}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[\\log_{}\\left|2x+{\\sqrt[]{4x^{2}+1}}\\right|-\\log_{}\\left|2\\right|\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left|2x+{\\sqrt[]{4x^{2}+1}}\\right|+c}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-1-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1600681960135,"cs":"i1hg4vf2u/j0kdT53DZu7A==","size":{"width":345,"height":268}}$

${"code":"$3.\\,\\frac{1}{{\\sqrt[]{\\left(2-x\\right)^{2}+1}}}$","type":"$","id":"1-0-1","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1600683112115,"cs":"i0taP9DMKBAGNffHrZTMIQ==","size":{"width":104,"height":40}}$

Solun:- Let f(x) = ${"type":"$","font":{"color":"#000000","family":"Arial","size":10},"id":"5-1","code":"$\\frac{1}{{\\sqrt[]{\\left(2-x\\right)^{2}+1}}}$","ts":1600683095117,"cs":"2GRdlgYYpVnTUHa9hRanjw==","size":{"width":66,"height":32}}$

Integrate f(x):-

${"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(2-x\\right)^{2}+1}}}.dx$","id":"2-0-0-0-0-0-0-1","type":"$","font":{"color":"#000000","family":"Arial","size":12},"ts":1600683072719,"cs":"XpBkcm1I9Zj/0bGESnHvwQ==","size":{"width":240,"height":40}}$

Let 2 - x = t

Differentiate w.r.t to t:-

⇒ - dx = dt

⇒ dx = - dt

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{-dt}{{\\sqrt[]{t^{2}+1}}}}\t\n\\end{align*}","id":"2-1-0-0-0-0-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1600683701074,"cs":"+Rmx4LSPkBmBkfEt9AiM/Q==","size":{"width":166,"height":37}}$

We know that:-

${"type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}+a^{2}}}\\right|+c$","id":"3-0-0-0-1-1-0","ts":1600681532828,"cs":"JsWT27ttO8B8cmfF/zeEJg==","size":{"width":257,"height":24}}$

${"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"2-1-1-0-0-0-0-0-0-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\log_{}\\left|t+{\\sqrt[]{t^{2}+1}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\frac{1}{t+{\\sqrt[]{t^{2}+1}}}\\right|+C}\t\n\\end{align*}","ts":1600683885561,"cs":"X2d4eTvbdB2Y3IUqO8743Q==","size":{"width":248,"height":80}}$

Put the value of t:-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\frac{1}{2-x+{\\sqrt[]{\\left(2-x\\right)^{2}+1}}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\frac{1}{2-x+{\\sqrt[]{x^{2}+4-4x+1}}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\frac{1}{2-x+{\\sqrt[]{x^{2}-4x+5}}}\\right|+C}\t\n\\end{align*}","font":{"size":8.461538461538462,"color":"#000000","family":"Arial"},"id":"2-1-1-0-0-1-0-0-1-0","type":"align*","ts":1600684117062,"cs":"AT28frXQyRz9qtaInGLO7g==","size":{"width":336,"height":132}}$

${"id":"1-1-1-0","code":"$4.\\,\\frac{1}{{\\sqrt[]{9-25x^{2}}}}$","font":{"family":"Arial","size":12,"color":"#000000"},"type":"$","ts":1600684422357,"cs":"xA6eoDw2ZnHe4G0RrL7eaQ==","size":{"width":86,"height":29}}$

Solun:- Let f(x) = ${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\frac{1}{{\\sqrt[]{9-25x^{2}}}}$","id":"6-1-0","type":"$","ts":1600684445886,"cs":"7iLXUAKP4BqxE3HHbph+1w==","size":{"width":52,"height":22}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{9-25x^{2}}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{25\\left(\\frac{9}{25}-x^{2}\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{5}\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(\\frac{9}{25}-x^{2}\\right)}}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1600684526977,"cs":"q9mxaKcduohSkqbVFIJs2A==","size":{"width":240,"height":148}}$

We know that:-

${"font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{a^{2}-x^{2}}}}.dx=\\sin^{-1}\\left(\\frac{x}{a}\\right)+c$","id":"3-0-0-0-1-0-1-0","type":"$","ts":1600687173678,"cs":"cYU2wGJmI2+KbsP55EiL0A==","size":{"width":196,"height":22}}$

${"type":"align*","id":"2-1-1-0-0-0-0-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{5}\\sin^{-1}\\left(\\frac{x}{\\frac{3}{5}}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{5}\\sin^{-1}\\left(\\frac{5x}{3}\\right)+C}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600687295054,"cs":"x72VH7jCCcovNhM533FD+A==","size":{"width":218,"height":89}}$

${"code":"$5.\\,\\frac{3x}{1+2x^{4}}$","id":"1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1600687377773,"cs":"8xMQN4+Ke0lBn4quorfM9A==","size":{"width":68,"height":26}}$

Solun:- Let f(x) = ${"type":"$","id":"6-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\frac{3x}{1+2x^{4}}$","ts":1600687411934,"cs":"XsNC9Mioyww63OnlKkUGPQ==","size":{"width":37,"height":20}}$

Integrate f(x):-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{3x}{1+2x^{4}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{3x}{1+2\\left(x^{2}\\right)^{2}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600687470248,"cs":"X5HUxMRH9jtnNO0uVL+izw==","size":{"width":208,"height":80}}$

Let x2 = t

Differentiate w.r.t to t:-

⇒ 2x.dx = dt

⇒ x.dx = (1/2).dt

${"id":"2-1-0-0-0-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{3}{2}\\int_{}^{}\\frac{1}{2\\left(\\frac{1}{2}+t^{2}\\right)}.dt}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{3}{4}\\int_{}^{}\\frac{1}{\\left(\\frac{1}{2}+t^{2}\\right)}.dt}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1600688171919,"cs":"N210iAE+nunGTq2fezfgTA==","size":{"width":218,"height":84}}$

We know that:-

${"id":"3-0-0-0-1-1-1-0","type":"$","code":"$\\int_{}^{}\\frac{1}{x^{2}+a^{2}}.dx=\\frac{1}{a}.\\tan^{-1}\\left(\\frac{x}{a}\\right)+c$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600688543272,"cs":"4QS7RR+zJzVW5r2hhdnaWA==","size":{"width":210,"height":20}}$

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{3}{4}.\\frac{1}{\\frac{1}{{\\sqrt[]{2}}}}.\\tan^{-1}\\left(\\frac{t}{\\frac{1}{{\\sqrt[]{2}}}}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{3{\\sqrt[]{2}}}{4}\\tan^{-1}\\left(t{\\sqrt[]{2}}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{3}{2{\\sqrt[]{2}}}\\tan^{-1}\\left(t{\\sqrt[]{2}}\\right)+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600688982908,"cs":"qhQcpJulKAX58BJl+Ql4nA==","size":{"width":274,"height":140}}$

Put the value of t:-

${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{3}{2{\\sqrt[]{2}}}\\tan^{-1}\\left({\\sqrt[]{2}}x^{2}\\right)+C}\t\n\\end{align*}","type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-1-0","ts":1600689095836,"cs":"c+gblfI3xjECpxCgs88cUw==","size":{"width":252,"height":37}}$

${"id":"1-1-1-1-1-0","code":"$6.\\,\\frac{x^{2}}{1-x^{6}}$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1600689187469,"cs":"kzOo4d+K4EtwxeWSFxC12A==","size":{"width":60,"height":28}}$

Solun:- Let f(x) = ${"font":{"color":"#000000","family":"Arial","size":10},"id":"6-1-1-1-0","type":"$","code":"$\\frac{x^{2}}{1-x^{6}}$","ts":1600689204145,"cs":"OduPiYyyi3EqXlNLNs88lg==","size":{"width":32,"height":21}}$

Integrate f(x):-

${"type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{2}}{1-x^{6}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{2}}{1-\\left(x^{3}\\right)^{2}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-1-1-0","ts":1600689257425,"cs":"mozAxzsg5imQumhbwgZnqQ==","size":{"width":200,"height":84}}$

Let x3 = t

Differentiate w.r.t to t:-

⇒ 3x2.dx = dt

⇒ x2.dx = (1/3).dt

${"font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-1-0-0-0-0-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{1-t^{2}}.\\frac{dt}{3}}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{3}\\int_{}^{}\\frac{1}{1-t^{2}}.dt}\t\n\\end{align*}","type":"align*","ts":1600689668538,"cs":"JCiXuf48qY/CXDl4t+Fd2g==","size":{"width":192,"height":76}}$

We know that:-

${"id":"3-0-0-0-1-1-1-1-0","code":"$\\int_{}^{}\\frac{1}{a^{2}-x^{2}}.dx=\\frac{1}{2a}.\\log_{}\\left|\\frac{a+x}{a-x}\\right|+c$","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","ts":1600689753460,"cs":"R8YiDflNtL66FmmG/G+M5g==","size":{"width":209,"height":20}}$

${"type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{3}.\\frac{1}{2}.\\log_{}\\left|\\frac{1+t}{1-t}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{6}\\log_{}\\left|\\frac{1+t}{1-t}\\right|+C}\t\n\\end{align*}","ts":1600689827729,"cs":"qR5qtD7aDu0zIv+h5wPA3A==","size":{"width":236,"height":76}}$

Put the value of t:-

${"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{6}\\log_{}\\left|\\frac{1+x^{3}}{1-x^{3}}\\right|+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-1-1-0","type":"align*","ts":1600689872112,"cs":"F4Fog7OSaQvG04a5wgEOPw==","size":{"width":217,"height":40}}$

${"code":"$7.\\,\\frac{x-1}{{\\sqrt[]{x^{2}-1}}}$","id":"1-1-1-1-1-1-0","type":"$","font":{"family":"Arial","color":"#000000","size":12},"ts":1600692474140,"cs":"2CwAJZcXfAB9vSR9OoXIHg==","size":{"width":72,"height":29}}$

Solun:- Let f(x) = ${"type":"$","code":"$\\frac{x-1}{{\\sqrt[]{x^{2}-1}}}$","id":"6-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1600692535267,"cs":"0C5CZRYZMczTCdbXJ0G6Tg==","size":{"width":41,"height":22}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-1-1-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x-1}{{\\sqrt[]{x^{2}-1}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x}{{\\sqrt[]{x^{2}-1}}}.dx-\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-1}}}.dx}\t\n\\end{align*}","ts":1600692599343,"cs":"AcXLl3dD1ewWwD5bqAXPKA==","size":{"width":320,"height":80}}$

Let x2 - 1 = t2

Differentiate w.r.t to t:-

⇒ 2x.dx = 2t.dt

⇒ x.dx = t.dt

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{t}{t}.dt-\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-1}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}1.dt-\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-1}}}.dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"2-1-0-0-0-0-1-1-1-1-0","ts":1600693195358,"cs":"xmjkvgvJSqImrMaRmXhfGQ==","size":{"width":268,"height":80}}$

We know that:-

${"font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}-a^{2}}}\\right|+c$","id":"3-0-0-0-1-1-1-1-1-0-0","ts":1600693366667,"cs":"+ZZ7dZuNkFK+3FhqbrbXdA==","size":{"width":257,"height":24}}$

${"id":"3-0-0-0-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"code":"$\\int_{}^{}1.dx=x+c$","type":"$","ts":1600693392542,"cs":"fWJAg6z5SP2UaxiKoWeS+A==","size":{"width":100,"height":17}}$

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={t-\\log_{}\\left|x+{\\sqrt[]{x^{2}-1}}\\right|+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1600693433219,"cs":"76MsnK2Dt9WGl+RlaV0Qgg==","size":{"width":265,"height":36}}$

Put the value of t:-

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-1","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={{\\sqrt[]{x^{2}-1}}-\\log_{}\\left|x+{\\sqrt[]{x^{2}-1}}\\right|+C}\t\n\\end{align*}","type":"align*","ts":1600693593729,"cs":"Kiq/OVI1XJzazdoj77qJvg==","size":{"width":318,"height":36}}$

${"id":"1-1-1-1-1-1-1","code":"$8.\\,\\frac{x^{2}}{{\\sqrt[]{x^{6}+a^{6}}}}$","font":{"color":"#000000","size":12,"family":"Arial"},"type":"$","ts":1600760648537,"cs":"tb5PBNiFXdv5SwZDZ5Qspg==","size":{"width":78,"height":32}}$

Solun:- Let f(x) = ${"code":"$\\frac{x^{2}}{{\\sqrt[]{x^{6}+a^{6}}}}$","id":"6-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"type":"$","ts":1600760694242,"cs":"eiWkeaIKnkfHhvMY5534eg==","size":{"width":46,"height":24}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-1-1-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{2}}{{\\sqrt[]{x^{6}+a^{6}}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{2}}{{\\sqrt[]{\\left(x^{3}\\right)^{2}+\\left(a^{3}\\right)^{2}}}}.dx}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600760749247,"cs":"TAEvfhw/keGUWaU5aGlpBw==","size":{"width":241,"height":97}}$

Let x3 = t

Differentiate w.r.t to t:-

⇒ 3x2.dx = dt

⇒ x2.dx = (1/3).dt

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{t^{2}+\\left(a^{3}\\right)^{2}}}}.\\frac{dt}{3}}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{3}\\int_{}^{}\\frac{1}{{\\sqrt[]{t^{2}+\\left(a^{3}\\right)^{2}}}}.dt}\t\n\\end{align*}","type":"align*","id":"2-1-0-0-0-0-1-1-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600760932106,"cs":"rRtieyw3dwcDIQu06lMKMg==","size":{"width":233,"height":105}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}+a^{2}}}\\right|+c$","id":"3-0-0-0-1-1-1-1-1-0-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","ts":1600760968849,"cs":"Vr22J6eWOW0Z9+rl95BIYg==","size":{"width":257,"height":24}}$

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{3}\\log_{}\\left|t+{\\sqrt[]{t^{2}+\\left(a^{3}\\right)^{2}}}\\right|+C}\t\n\\end{align*}","type":"align*","ts":1600761079907,"cs":"lCycJrmmArVte9KHpoiLUw==","size":{"width":274,"height":36}}$

Put the value of t:-

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-1-0","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{3}\\log_{}\\left|x^{3}+{\\sqrt[]{\\left(x^{3}\\right)^{2}+\\left(a^{3}\\right)^{2}}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{3}\\log_{}\\left|x^{3}+{\\sqrt[]{x^{6}+a^{6}}}\\right|+C}\t\n\\end{align*}","type":"align*","ts":1600761203830,"cs":"3Rq6L4L8APQq21Q4XcwjPw==","size":{"width":305,"height":77}}$

${"type":"$","code":"$9.\\,\\frac{\\sec^{2}x}{{\\sqrt[]{\\tan^{2}x+4}}}$","font":{"color":"#000000","family":"Arial","size":12},"id":"1-1-1-1-1-1-2-0","ts":1600761684572,"cs":"+Cd8xeNL3a+11edvQ7D3vw==","size":{"width":98,"height":34}}$

Solun:- Let f(x) = ${"id":"6-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"type":"$","code":"$\\frac{\\sec^{2}x}{{\\sqrt[]{\\tan^{2}x+4}}}$","ts":1600761702679,"cs":"q9UzKlW2r5miD4AR9l56IQ==","size":{"width":61,"height":26}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\sec^{2}x}{{\\sqrt[]{\\tan^{2}x+4}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\sec^{2}x}{{\\sqrt[]{\\tan^{2}x+2^{2}}}}.dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1600761738315,"cs":"hZp7rb5CfUweEHI5SwoGcQ==","size":{"width":229,"height":90}}$

Let tan x = t

Differentiate w.r.t to t:-

⇒ sec2x.dx = dt

${"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{{\\sqrt[]{t^{2}+\\left(2\\right)^{2}}}}}\t\n\\end{align*}","id":"2-1-0-0-0-0-1-1-1-1-1-1-0","type":"align*","ts":1600761895120,"cs":"OURNBFKkNd823NYmvD9/sA==","size":{"width":188,"height":50}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}+a^{2}}}\\right|+c$","id":"3-0-0-0-1-1-1-1-1-0-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","ts":1600760968849,"cs":"nuP2MQh0x6A9rXHjS83veA==","size":{"width":257,"height":24}}$

${"type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|t+{\\sqrt[]{t^{2}+2^{2}}}\\right|+C}\t\n\\end{align*}","ts":1600761975729,"cs":"wLFUpLK6nOIpGAQDYar/PQ==","size":{"width":240,"height":36}}$

Put the value of t:-

${"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\tan x+{\\sqrt[]{\\tan^{2}x+4}}\\right|+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-1-1-0","type":"align*","ts":1600762075594,"cs":"kTdssTlsxyfqpsdfnG9lHQ==","size":{"width":292,"height":36}}$

${"id":"1-1-1-1-1-1-2-1-0","type":"$","code":"$10.\\,\\frac{1}{{\\sqrt[]{x^{2}+2x+2}}}$","font":{"color":"#000000","size":12,"family":"Arial"},"ts":1600770812168,"cs":"qJDR7XqOm/StXnT2MwYwYQ==","size":{"width":109,"height":29}}$

Solun:- Let f(x) = ${"font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","id":"6-1-1-1-1-1-1-1-0","code":"$\\frac{1}{{\\sqrt[]{x^{2}+2x+2}}}$","ts":1600770859247,"cs":"7mhzP7bGQJ4MhcpX3PNYcw==","size":{"width":61,"height":22}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+2x+2}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+2x+\\left(1\\right)^{2}-\\left(1\\right)^{2}+2}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x+1\\right)^{2}+1}}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-0","type":"align*","font":{"family":"Arial","color":"#000000","size":5.841269841269841},"ts":1600771162664,"cs":"LaIJmtFKu0qN6UYWDWi6mw==","size":{"width":278,"height":101}}$

Let x + 1 = t

Differentiate w.r.t to t:-

⇒ dx = dt

${"type":"align*","id":"2-1-0-0-0-0-1-1-1-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{{\\sqrt[]{t^{2}+\\left(1\\right)^{2}}}}}\t\n\\end{align*}","ts":1600771223702,"cs":"nyUiElsQpF8OaQ9WTDfRHQ==","size":{"width":188,"height":50}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}+a^{2}}}\\right|+c$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","ts":1600760968849,"cs":"VVCff67Ym179kAuFm9meYA==","size":{"width":257,"height":24}}$

${"type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|t+{\\sqrt[]{t^{2}+1}}\\right|+C}\t\n\\end{align*}","ts":1600771280699,"cs":"9ShqtlFsuP8Pe8VYMIwq4A==","size":{"width":233,"height":36}}$

Put the value of t:-

${"font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|x+1+{\\sqrt[]{\\left(x+1\\right)^{2}+1}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|x+1+{\\sqrt[]{x^{2}+2x+2}}\\right|+C}\t\n\\end{align*}","type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-1-1-1-0","ts":1600771370562,"cs":"+QRQLbmuc9hYc52hKN4ImQ==","size":{"width":308,"height":77}}$

${"type":"$","font":{"color":"#000000","family":"Arial","size":12},"id":"1-1-1-1-1-1-2-1-1-0","code":"$11.\\,\\frac{1}{9x^{2}+6x+5}$","ts":1600771733608,"cs":"n3yfxe/6OZKUG+r1BZqIUQ==","size":{"width":104,"height":25}}$

Solun:- Let f(x) = ${"font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\frac{1}{9x^{2}+6x+5}$","id":"6-1-1-1-1-1-1-1-1-0-0","ts":1600771757240,"cs":"RGtPuYfFFYJ5MJRopTOMNg==","size":{"width":58,"height":20}}$

Integrate f(x):-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{9x^{2}+6x+5}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{9\\left(x^{2}+\\frac{6}{9}x+\\frac{5}{9}\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{9}\\int_{}^{}\\frac{1}{\\left(x^{2}+\\frac{2}{3}x+\\left(\\frac{1}{3}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}+\\frac{5}{9}\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{9}\\int_{}^{}\\frac{1}{\\left(\\left(x+\\frac{1}{3}\\right)^{2}-\\frac{1}{9}+\\frac{5}{9}\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{9}\\int_{}^{}\\frac{1}{\\left(\\left(x+\\frac{1}{3}\\right)^{2}+\\frac{4}{9}\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{9}\\int_{}^{}\\frac{1}{\\left(\\left(x+\\frac{1}{3}\\right)^{2}+\\left(\\frac{2}{3}\\right)^{2}\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-0-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1600771975760,"cs":"2vIHxzaUPhFE9GT/HA0CVQ==","size":{"width":368,"height":296}}$

Let x + 1/3 = t

Differentiate w.r.t to t:-

⇒ dx = dt

${"id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{9}\\int_{}^{}\\frac{dt}{t^{2}+\\left(\\frac{2}{3}\\right)^{2}}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601033016713,"cs":"It/fV0HnR9Rf/qcbpJHxfQ==","size":{"width":196,"height":44}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{x^{2}+a^{2}}.dx=\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+c$","font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-0-0","ts":1600772250107,"cs":"gksdS5i2utqAlxi/AC0Rlw==","size":{"width":204,"height":20}}$

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-0-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{9}\\times\\frac{1}{\\frac{2}{3}}\\tan^{-1}\\left(\\frac{t}{\\frac{2}{3}}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{9}\\times\\frac{3}{2}\\tan^{-1}\\left(\\frac{3t}{2}\\right)+C}\t\n\\end{align*}","ts":1600772533577,"cs":"Zv6GioI5KHYAgRl2XPV8rQ==","size":{"width":257,"height":89}}$

Put the value of t:-

${"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{6}\\tan^{-1}\\left(\\frac{3\\left(x+\\frac{1}{3}\\right)}{2}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{6}\\tan^{-1}\\left(\\frac{3x+1}{2}\\right)+C}\t\n\\end{align*}","ts":1600772556598,"cs":"VJMNIAaYvwwT4/AIgKw9XQ==","size":{"width":269,"height":89}}$

${"type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"code":"$12.\\,\\frac{1}{{\\sqrt[]{7-6x-x^{2}}}}$","id":"1-1-1-1-1-1-2-1-1-1-0","ts":1600772635181,"cs":"jhUJnlz62DNQwGe2MHZkqQ==","size":{"width":109,"height":29}}$

Solun:- Let f(x) = ${"code":"$\\frac{1}{{\\sqrt[]{7-6x-x^{2}}}}$","id":"6-1-1-1-1-1-1-1-1-1-0","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1600772707023,"cs":"Xjh6Pv8q0GousazebzHN5w==","size":{"width":61,"height":22}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{7-6x-x^{2}}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{-\\left(x^{2}+6x-7\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{-\\left(x^{2}+6x+\\left(3\\right)^{2}-\\left(3\\right)^{2}-7\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{-\\left(\\left(x+3\\right)^{2}-16\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{16-\\left(x+3\\right)^{2}}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(4\\right)^{2}-\\left(x+3\\right)^{2}}}}.dx}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0","ts":1600772969686,"cs":"35BXK0T0Of2rhIwAnM6MAg==","size":{"width":356,"height":322}}$

Let x + 3 = t

Differentiate w.r.t to t:-

⇒ dx = dt

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{{\\sqrt[]{\\left(4\\right)^{2}-t^{2}}}}}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-0","ts":1600773087852,"cs":"927v9XV0iLLSZiwWCizYbw==","size":{"width":188,"height":50}}$

We know that:-

${"type":"$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-0","code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{a^{2}-x^{2}}}}.dx=\\sin^{-1}\\left(\\frac{x}{a}\\right)+c$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1600773157216,"cs":"miiP7WRIfXEZ6v/XtGKjBg==","size":{"width":196,"height":22}}$

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\sin^{-1}\\left(\\frac{t}{4}\\right)+C}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1600773288617,"cs":"hgwypsH5ICzrjK28JhUHyA==","size":{"width":194,"height":37}}$

Put the value of t:-

${"font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\sin^{-1}\\left(\\frac{x+3}{4}\\right)+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-1-1-0","type":"align*","ts":1600773271965,"cs":"HClJm8SjshanGaLltVdTOQ==","size":{"width":222,"height":37}}$

${"type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"code":"$13.\\,\\frac{1}{{\\sqrt[]{\\left(x-1\\right)\\left(x-2\\right)}}}$","id":"1-1-1-1-1-1-2-1-1-1-1-0","ts":1600773370112,"cs":"nff5c3OKlKZ5QxUe1TTy6w==","size":{"width":128,"height":32}}$

Solun:- Let f(x) = ${"font":{"family":"Arial","size":10,"color":"#000000"},"id":"6-1-1-1-1-1-1-1-1-1-1-0","type":"$","code":"$\\frac{1}{{\\sqrt[]{\\left(x-1\\right)\\left(x-2\\right)}}}$","ts":1600773387522,"cs":"gbaFGfUiFWIez+a16wUVbA==","size":{"width":76,"height":24}}$

Integrate f(x):-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x-1\\right)\\left(x-2\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-3x+2}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-3x+\\left(\\frac{3}{2}\\right)^{2}-\\left(\\frac{3}{2}\\right)^{2}+2}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x-\\frac{3}{2}\\right)^{2}-\\frac{1}{4}}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x-\\frac{3}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}}}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1600774051755,"cs":"5YlqGkMmWTfSOmLC1RYE5Q==","size":{"width":340,"height":248}}$

Let x - 3/2 = t

Differentiate w.r.t to t:-

⇒ dx = dt

${"id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{{\\sqrt[]{t^{2}-\\left(\\frac{1}{2}\\right)^{2}}}}}\t\n\\end{align*}","ts":1600773717624,"cs":"tPnas0jX2puwzPSvVFJ4rA==","size":{"width":196,"height":50}}$

We know that:-

${"id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-0","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}-a^{2}}}\\right|+c$","ts":1600773813551,"cs":"hZm58JXqcnGSEIg8RGa8Ng==","size":{"width":257,"height":24}}$

${"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|t+{\\sqrt[]{t^{2}-{\\left(\\frac{1}{2}\\right)}^{2}}}\\right|+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-0","ts":1600773863856,"cs":"EJiTB9Ns+NJqOYqgh7KMoA==","size":{"width":270,"height":56}}$

Put the value of t:-

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\left(x-\\frac{3}{2}\\right)+{\\sqrt[]{\\left(x-\\frac{3}{2}\\right)^{2}-{\\left(\\frac{1}{2}\\right)}^{2}}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\left(x-\\frac{3}{2}\\right)+{\\sqrt[]{x^{2}-3x+2}}\\right|+C}\t\n\\end{align*}","type":"align*","ts":1601033361719,"cs":"3LOW0On69dB8dWYqcPGKoQ==","size":{"width":392,"height":98}}$

${"font":{"family":"Arial","size":12,"color":"#000000"},"id":"1-1-1-1-1-1-2-1-1-1-1-1-0","type":"$","code":"$14.\\,\\frac{1}{{\\sqrt[]{8+3x-x^{2}}}}$","ts":1600774667628,"cs":"HtcXntPvWH0zLwgeP3XhTg==","size":{"width":109,"height":29}}$

Solun:- Let f(x) = ${"font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\frac{1}{{\\sqrt[]{8+3x-x^{2}}}}$","id":"6-1-1-1-1-1-1-1-1-1-1-1-0","ts":1600774720044,"cs":"GYBPwr6O9zzo2kA1vvdDUQ==","size":{"width":61,"height":22}}$

Integrate f(x):-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{8+3x-x^{2}}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{-\\left(x^{2}-3x-8\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{-\\left(x^{2}-3x+\\left(\\frac{3}{2}\\right)^{2}-\\left(\\frac{3}{2}\\right)^{2}-8\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{-\\left(\\left(x-\\frac{3}{2}\\right)^{2}-\\frac{41}{4}\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(\\frac{{\\sqrt[]{41}}}{2}\\right)^{2}-\\left(x-\\frac{3}{2}\\right)^{2}}}}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1600775157397,"cs":"Qr4kwAiewGjoniEogmtHPQ==","size":{"width":370,"height":276}}$

Let x - 3/2 = t

Differentiate w.r.t to t:-

⇒ dx = dt

${"font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{{\\sqrt[]{\\left(\\frac{{\\sqrt[]{41}}}{2}\\right)^{2}-t^{2}}}}}\t\n\\end{align*}","type":"align*","ts":1600775402172,"cs":"1MH9WcVrHh0UqQFudHY7lg==","size":{"width":214,"height":60}}$

We know that:-

${"font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{a^{2}-x^{2}}}}.dx=\\sin^{-1}\\left(\\frac{x}{a}\\right)+c$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-0","type":"$","ts":1600775477454,"cs":"L6qP63HAp0ZYEDKkayZ1hw==","size":{"width":196,"height":22}}$

${"type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\sin^{-1}\\left(\\frac{t}{\\frac{{\\sqrt[]{41}}}{2}}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\sin^{-1}\\left(\\frac{2t}{{\\sqrt[]{41}}}\\right)+C}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1600775532636,"cs":"P0h0TJlxMcMW9HK8dW1JoA==","size":{"width":216,"height":93}}$

Put the value of t:-

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\sin^{-1}\\left(\\frac{2\\left(x-\\frac{3}{2}\\right)}{{\\sqrt[]{41}}}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\sin^{-1}\\left(\\frac{2x-3}{{\\sqrt[]{41}}}\\right)+C}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1600775611226,"cs":"dtG53NzKBluLEkErA93sXw==","size":{"width":252,"height":90}}$

${"type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-0","code":"$15.\\,\\frac{1}{{\\sqrt[]{\\left(x-a\\right)\\left(x-b\\right)}}}$","ts":1600848398702,"cs":"lVXbcr/8VM4voEPEzTUarw==","size":{"width":128,"height":32}}$

Solun:- Let f(x) = ${"code":"$\\frac{1}{{\\sqrt[]{\\left(x-a\\right)\\left(x-b\\right)}}}$","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"6-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1600848416239,"cs":"L8FIhrhoFqTQ4niqRxkGSg==","size":{"width":76,"height":24}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x-a\\right)\\left(x-b\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-x\\left(a+b\\right)+ab}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-x\\left(a+b\\right)+\\left(\\frac{a+b}{2}\\right)^{2}-\\left(\\frac{a+b}{2}\\right)^{2}+ab}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x-\\frac{\\left(a+b\\right)}{2}\\right)^{2}-\\frac{\\left(a-b\\right)^{2}}{4}}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x-\\frac{\\left(a+b\\right)}{2}\\right)^{2}-\\left[\\frac{\\left(a-b\\right)}{2}\\right]^{2}}}}.dx}\t\n\\end{align*}","ts":1600848988727,"cs":"G5wf1whL3GAjd7u/51NRjA==","size":{"width":413,"height":270}}$

Let x - (a+b)/2 = t

Differentiate w.r.t to t:-

⇒ dx = dt

${"font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{{\\sqrt[]{t^{2}-{\\left[\\frac{a-b}{2}\\right]}^{2}}}}}\t\n\\end{align*}","type":"align*","ts":1600849205028,"cs":"8X5552xQGdCyWQTYgum5oQ==","size":{"width":208,"height":50}}$

We know that:-

${"font":{"color":"#000000","family":"Arial","size":10},"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}-a^{2}}}\\right|+c$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-0","type":"$","ts":1600849308995,"cs":"6FPFRARWpNOnrd3c/0azqg==","size":{"width":257,"height":24}}$

${"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|t+{\\sqrt[]{t^{2}-\\left(\\frac{a-b}{2}\\right)^{2}}}\\right|+C}\t\n\\end{align*}","ts":1600849419089,"cs":"C9jclZcN7+A5w2IrdsQNeg==","size":{"width":297,"height":56}}$

Put the value of t:-

${"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|x-\\frac{\\left(a+b\\right)}{2}+{\\sqrt[]{{\\left[x-\\frac{\\left(a+b\\right)}{2}\\right]}^{2}-\\left(\\frac{a-b}{2}\\right)^{2}}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|x-\\frac{\\left(a+b\\right)}{2}+{\\sqrt[]{\\left(x-a\\right)\\left(x-b\\right)}}\\right|+C}\t\n\\end{align*}","ts":1600858134953,"cs":"tCNfx+UWG6++EdZMqv+/Ug==","size":{"width":466,"height":98}}$

${"code":"$16.\\,\\frac{4x+1}{{\\sqrt[]{2x^{2}+x-3}}}$","id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-0","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1600863321920,"cs":"lbV4IFAKVzpSTUvMPfNIKw==","size":{"width":109,"height":29}}$

Solun:- Let f(x) = ${"code":"$\\frac{4x+1}{{\\sqrt[]{2x^{2}+x-3}}}$","type":"$","id":"6-1-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1600863342686,"cs":"7z0AlpxUEOJXYorowHBNEQ==","size":{"width":61,"height":22}}$

Integrate f(x):-

${"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{4x+1}{{\\sqrt[]{2x^{2}+x-3}}}.dx}\t\n\\end{align*}","ts":1600863699269,"cs":"onHPxTPM9e51kF7BkzQlpQ==","size":{"width":230,"height":37}}$

Let 2x2 + x - 3 = t2

Differentiate w.r.t to t:-

⇒ (4x + 1).dx = 2t.dt

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{2t.dt}{t}}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={2\\int_{}^{}1.dt}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1600865110013,"cs":"HbieLnnfWtx87pu6wqzM+A==","size":{"width":148,"height":76}}$

We know that:-

${"font":{"color":"#000000","family":"Arial","size":10},"code":"$\\int_{}^{}1.dx=x+c$","type":"$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-0","ts":1600865133164,"cs":"4H+JCsSnEsLYTI1HRBs4qQ==","size":{"width":100,"height":17}}$

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={2t+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-0","ts":1600865151454,"cs":"ZXkH1y16XOAueWvUErK+GQ==","size":{"width":132,"height":36}}$

Put the value of t:-

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={2{\\sqrt[]{2x^{2}+x-3}}+C}\t\n\\end{align*}","ts":1600865214410,"cs":"vUgwIQ4sPKtz37kf5plgEg==","size":{"width":222,"height":36}}$

${"font":{"size":12,"family":"Arial","color":"#000000"},"type":"$","code":"$17.\\,\\frac{x+2}{{\\sqrt[]{x^{2}-1}}}$","id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0","ts":1600865328985,"cs":"48qkg8YIs+YBPJDJdyqKkQ==","size":{"width":82,"height":29}}$

Solun:- Let f(x) = ${"id":"6-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","code":"$\\frac{x+2}{{\\sqrt[]{x^{2}-1}}}$","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1600865427002,"cs":"KJJzT34eGqqwnjIMf7vpkQ==","size":{"width":41,"height":22}}$

Integrate f(x):-

${"font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x+2}{{\\sqrt[]{x^{2}-1}}}.dx}\t\n\\end{align*}","ts":1600865444145,"cs":"22V/meMxLd9uvY4iYwmsdw==","size":{"width":194,"height":37}}$

Using Formula:-

Numerator = A.d(Denominator)/dx + B

${"code":"\\begin{align*}\n{x+2}&={A\\diff{\\left(x^{2}-1\\right)}{x}+B}\\\\\n{x+2}&={A\\left(2x\\right)+B...\\left(1\\right)}\\\\\n{x+2}&={2Ax+B}\t\n\\end{align*}","id":"7-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600865512999,"cs":"TJSqlBaKWhRdXDjcKR1X8w==","size":{"width":173,"height":77}}$

Compare both side:-

⇒ 2A = 1 then A = 1/2

⇒ B = 2

From eq. 1:-

⇒ x + 2 = (½).(2x)+2

${"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\frac{1}{2}\\left(2x\\right)+2}{{\\sqrt[]{x^{2}-1}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\frac{\\left(2x\\right)}{{\\sqrt[]{x^{2}-1}}}.dx+2\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-1}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}I_{1}+2I_{2}...\\left(2\\right)}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1600865928660,"cs":"21gPdjhlE2xm/DU9Yfpsvw==","size":{"width":348,"height":126}}$

Calculate I1:-

${"code":"$I_{1}=\\int_{}^{}\\frac{\\left(2x\\right)}{{\\sqrt[]{x^{2}-1}}}.dx$","type":"$","id":"8-0-0","font":{"size":12,"color":"#222222","family":"Arial"},"ts":1600865978669,"cs":"xXD1zIvxWcSafi2Fvk9FAg==","size":{"width":146,"height":32}}$

Let x2 - 1 = t2

Differentiate w.r.t to t:-

⇒ 2x.dx = 2t.dt

${"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{I_{1}}&={\\int_{}^{}\\frac{2t.dt}{t}}\\\\\n{I_{1}}&={2\\int_{}^{}1.dt}\t\n\\end{align*}","id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1600866290589,"cs":"3n2e0MzF3eFxjoOUCJCuaA==","size":{"width":96,"height":76}}$

We know that:-

${"font":{"color":"#000000","family":"Arial","size":10},"code":"$\\int_{}^{}1.dx=x+c$","type":"$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-0","ts":1600865133164,"cs":"vL6tsc/wzLZ85SLs91GRzg==","size":{"width":100,"height":17}}$

${"type":"align*","code":"\\begin{align*}\n{I_{1}}&={2t+C}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-0-0","ts":1600866323415,"cs":"5OdkaPa1kvNGkcWiMZte2g==","size":{"width":80,"height":16}}$

Put the value of t:-

${"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{I_{1}}&={2{\\sqrt[]{x^{2}-1}}+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-1-1-1-0","ts":1600866345827,"cs":"xifDKlwykSfY1ICNN8+eZg==","size":{"width":132,"height":21}}$

Calculate I2:-

${"code":"$I_{2}=\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-1}}}.dx$","id":"8-1-0","type":"$","font":{"family":"Arial","color":"#222222","size":12},"ts":1600866406406,"cs":"L9a/MNs/mAlJymrpmtCWtQ==","size":{"width":146,"height":29}}$

We know that:-

${"id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-2-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}-a^{2}}}\\right|+c$","type":"$","ts":1600866513215,"cs":"2cn+nEsYCAZ5g2gVNCdi9g==","size":{"width":257,"height":24}}$

${"font":{"family":"Arial","color":"#000000","size":10},"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-1-0","type":"align*","code":"\\begin{align*}\n{I_{2}}&={\\log_{}\\left|x+{\\sqrt[]{x^{2}-1}}\\right|+C}\t\n\\end{align*}","ts":1600866553356,"cs":"k9Ng2T8cc4/rP3kaXsq6gw==","size":{"width":186,"height":26}}$

Put the value of I1 and I2 in Eq. 2:-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}.2{\\sqrt[]{x^{2}-1}}+2\\log_{}\\left|x+{\\sqrt[]{x^{2}-1}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={{\\sqrt[]{x^{2}-1}}+2\\log_{}\\left|x+{\\sqrt[]{x^{2}-1}}\\right|+C}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1","font":{"color":"#000000","family":"Arial","size":10},"ts":1601035041535,"cs":"e3ccsMzt+j+t5+t5FxBZzw==","size":{"width":356,"height":76}}$

${"type":"$","font":{"color":"#000000","family":"Arial","size":12},"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-0","code":"$18.\\,\\frac{5x-2}{1+2x+3x^{2}}$","ts":1600937528982,"cs":"nSFB3fD6Je9nwgMUNoxfOw==","size":{"width":104,"height":26}}$

Solun:- Let f(x) = ${"code":"$\\frac{5x-2}{1+2x+3x^{2}}$","type":"$","id":"6-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600937571020,"cs":"PTiu3JdcupFEu/zFn73Qtg==","size":{"width":58,"height":20}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{5x-2}{1+2x+3x^{2}}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1600937589318,"cs":"a/jktHtCUq4vk4IbGAnsDw==","size":{"width":225,"height":36}}$

Using Formula:-

Numerator = A.d(Denominator)/dx + B

${"font":{"family":"Arial","color":"#000000","size":10},"id":"7-1-0","code":"\\begin{align*}\n{5x-2}&={A\\diff{\\left(3x^{2}+2x+1\\right)}{x}+B}\\\\\n{5x-2}&={A\\left(6x+2\\right)+B...\\left(1\\right)}\\\\\n{5x-2}&={6Ax+2A+B}\t\n\\end{align*}","type":"align*","ts":1600937685479,"cs":"sCrf/o8AOvx3gylWQDqY+w==","size":{"width":225,"height":77}}$

Comare both side:-

⇒ 6A = 5 then A = 5/6

⇒ 2A + B = -2

⇒ 2.(⅚) + B = -2

⇒ B = -2 - 5/3

⇒ B = - 11/3

From eq. 1:-

⇒ 5x - 2 = (5/6).(6x+2)+(-11/3)

${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\frac{5}{6}\\left(6x+2\\right)-\\frac{11}{3}}{3x^{2}+2x+1}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{5}{6}\\int_{}^{}\\frac{\\left(6x+2\\right)}{3x^{2}+2x+1}.dx-\\frac{11}{3}\\int_{}^{}\\frac{1}{3x^{2}+2x+1}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{5}{6}I_{1}-\\frac{11}{3}I_{2}...\\left(2\\right)}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-2-0","type":"align*","ts":1600938236659,"cs":"6N96bty58QBViDIMK9JP2A==","size":{"width":425,"height":122}}$

Calculate I1:-

${"font":{"size":12,"color":"#222222","family":"Arial"},"type":"$","id":"8-0-1-0","code":"$I_{1}=\\int_{}^{}\\frac{\\left(6x+2\\right)}{3x^{2}+2x+1}.dx$","ts":1600938284220,"cs":"ONL3StW/idpOzHudIcGAiA==","size":{"width":168,"height":29}}$

Let 3x2 + 2x + 1 = t

Differentiate w.r.t to t:-

⇒ (6x + 2).dx = dt

${"code":"\\begin{align*}\n{I_{1}}&={\\int_{}^{}\\frac{dt}{t}}\t\n\\end{align*}","type":"align*","id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1600938358580,"cs":"wl1ZVsTehjuyy2SwW/LV7Q==","size":{"width":74,"height":36}}$

We know that:-

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-0","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","type":"$","ts":1600938463923,"cs":"u6MAqXx4S54GIZkkevLKng==","size":{"width":140,"height":18}}$

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-0-1-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I_{1}}&={\\log_{}\\left|t\\right|+C}\t\n\\end{align*}","ts":1600938489263,"cs":"+hB7t8aocCqd5wcf8U+kbQ==","size":{"width":104,"height":16}}$

Put the value of t:-

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-1-1-1-1-0","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{I_{1}}&={\\log_{}\\left|3x^{2}+2x+1\\right|+C}\t\n\\end{align*}","ts":1600938518518,"cs":"YmgzajiRon1XmLgDDMsvcA==","size":{"width":186,"height":20}}$

Calculate I2:-

${"id":"8-1-1-0","code":"\\begin{align*}\n{I_{2}}&={\\int_{}^{}\\frac{1}{3x^{2}+2x+1}.dx}\\\\\n{I_{2}}&={\\frac{1}{3}\\int_{}^{}\\frac{1}{x^{2}+\\frac{2}{3}x+\\left(\\frac{1}{3}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}+\\frac{1}{3}}.dx}\\\\\n{I_{2}}&={\\frac{1}{3}\\int_{}^{}\\frac{1}{\\left(x+\\frac{1}{3}\\right)^{2}+\\frac{2}{9}}.dx}\\\\\n{I_{2}}&={\\frac{1}{3}\\int_{}^{}\\frac{1}{\\left(x+\\frac{1}{3}\\right)^{2}+\\left(\\frac{{\\sqrt[]{2}}}{3}\\right)^{2}}.dx}\t\n\\end{align*}","font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","ts":1600938793720,"cs":"fBgW84diLJDOLFaImdoCwA==","size":{"width":296,"height":188}}$

We know that:-

${"type":"$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-2-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\int_{}^{}\\frac{1}{x^{2}+a^{2}}.dx=\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+c$","ts":1600938873695,"cs":"1+nBdEiKJZk6hb1SJMqacA==","size":{"width":204,"height":20}}$

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-1-1-0-0","code":"\\begin{align*}\n{I_{2}}&={\\frac{1}{3}\\times\\frac{1}{\\frac{{\\sqrt[]{2}}}{3}}\\tan^{-1}\\left(\\frac{x+\\frac{1}{3}}{\\frac{{\\sqrt[]{2}}}{3}}\\right)+C}\\\\\n{I_{2}}&={\\frac{1}{{\\sqrt[]{2}}}\\tan^{-1}\\left(\\frac{3\\left(x+\\frac{1}{3}\\right)}{{\\sqrt[]{2}}}\\right)+C}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1601034748159,"cs":"nIFzKYJ5hJrq9+9hkm3Y6w==","size":{"width":245,"height":104}}$

${"code":"\\begin{align*}\n{I_{2}}&={\\frac{1}{{\\sqrt[]{2}}}\\tan^{-1}\\left(\\frac{3x+1}{{\\sqrt[]{2}}}\\right)+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-1-1-0-1","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1601034789339,"cs":"ARWVadYKPCcHHchhxHuzVA==","size":{"width":208,"height":38}}$

Put the value of I1 and I2 in Eq. 2:-

${"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-3-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{5}{6}.\\log_{}\\left|3x^{2}+2x+1\\right|-\\frac{11}{3}\\times\\frac{1}{{\\sqrt[]{2}}}\\tan^{-1}\\left(\\frac{3x+1}{{\\sqrt[]{2}}}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{5}{6}.\\log_{}\\left|3x^{2}+2x+1\\right|-\\frac{11}{3{\\sqrt[]{2}}}\\tan^{-1}\\left(\\frac{3x+1}{{\\sqrt[]{2}}}\\right)+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601034916747,"cs":"lg/SETZuDYVKjGOw99tn+A==","size":{"width":468,"height":82}}$

${"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-1-0","type":"$","font":{"family":"Arial","color":"#000000","size":12},"code":"$19.\\,\\frac{6x+7}{{\\sqrt[]{\\left(x-5\\right)\\left(x-4\\right)}}}$","ts":1600943034494,"cs":"9d3+ZMCOp2qlHqBXIEjKWw==","size":{"width":128,"height":32}}$

Solun:- Let f(x) = ${"id":"6-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","code":"$\\frac{6x+7}{{\\sqrt[]{\\left(x-5\\right)\\left(x-4\\right)}}}$","font":{"color":"#000000","family":"Arial","size":10},"type":"$","ts":1600942589958,"cs":"AlnVlRwBzYdbCxo/x5bsLw==","size":{"width":76,"height":24}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{6x+7}{{\\sqrt[]{\\left(x-5\\right)\\left(x-4\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{6x+7}{{\\sqrt[]{x^{2}-9x+20}}}.dx}\\\\\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1600942607915,"cs":"zbhYeH+CdgazqG1P1wozoQ==","size":{"width":252,"height":82}}$

Using Formula:-

Numerator = A.d(Denominator)/dx + B

${"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"7-1-1-0","code":"\\begin{align*}\n{6x+7}&={A\\diff{\\left(x^{2}-9x+20\\right)}{x}+B}\\\\\n{6x+7}&={A\\left(2x-9\\right)+B...\\left(1\\right)}\\\\\n{6x+7}&={2Ax-9A+B}\t\n\\end{align*}","ts":1600942634030,"cs":"QyyyTjEIzz6KLkWxXEN0Tw==","size":{"width":225,"height":77}}$

Comare both side:-

⇒ 2A = 6 then A = 3

⇒ -9A + B = 7

⇒ -9.(3) + B = 7

⇒ B = 34

From eq. 1:-

⇒ 6x + 7 = (3).(2x - 9)+34

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{3\\left(2x-9\\right)+34}{{\\sqrt[]{x^{2}-9x+20}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={3\\int_{}^{}\\frac{\\left(2x-9\\right)}{{\\sqrt[]{x^{2}-9x+20}}}.dx+34\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-9x+20}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={3I_{1}+34I_{2}...\\left(2\\right)}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-2-1-0-0","ts":1600942936226,"cs":"Wctleh1k9ieE+ZGSA3CJ0w==","size":{"width":438,"height":124}}$

Calculate I1:-

${"code":"$I_{1}=\\int_{}^{}\\frac{\\left(2x-9\\right)}{{\\sqrt[]{x^{2}-9x+20}}}.dx$","id":"8-0-1-1-0-0","font":{"size":12,"color":"#222222","family":"Arial"},"type":"$","ts":1600941634179,"cs":"gOb/SzgoJuGoQyhbwIgskQ==","size":{"width":180,"height":32}}$

Let x2 - 9x + 20 = t2

Differentiate w.r.t to t:-

⇒ (2x - 9).dx = 2t.dt

${"id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{I_{1}}&={\\int_{}^{}\\frac{2t.dt}{t}}\\\\\n{I_{1}}&={2\\int_{}^{}1.dt}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1600941827577,"cs":"4bzJ0H2RiejQaHYsxqNYyg==","size":{"width":96,"height":76}}$

We know that:-

${"type":"$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-1-0","code":"$\\int_{}^{}1.dx=x+c$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600941859188,"cs":"C5C7oqcK70Amp3gbXyXoCQ==","size":{"width":100,"height":17}}$

${"type":"align*","code":"\\begin{align*}\n{I_{1}}&={2t+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-0-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1600941929003,"cs":"cJAWAv7GSXl4LknMQjTywA==","size":{"width":80,"height":16}}$

Put the value of t:-

${"type":"align*","code":"\\begin{align*}\n{I_{1}}&={2{\\sqrt[]{x^{2}-9x+20}}+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-1-1-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1600941950496,"cs":"wsIOb4ceFLqtXkjO5p3yGQ==","size":{"width":176,"height":21}}$

Calculate I2:-

${"type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"code":"\\begin{align*}\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-9x+20}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-9x+\\left(\\frac{9}{2}\\right)^{2}-\\left(\\frac{9}{2}\\right)^{2}+20}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x-\\frac{9}{2}\\right)^{2}+\\frac{1}{4}}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x-\\frac{9}{2}\\right)^{2}+\\left(\\frac{1}{2}\\right)^{2}}}}.dx}\t\n\\end{align*}","id":"8-1-1-1-0","ts":1601037596202,"cs":"Y3GpuYHpWOHyhtp5GBok5Q==","size":{"width":293,"height":202}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}+a^{2}}}\\right|+c$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-2-1-1-0","type":"$","ts":1600942248946,"cs":"uoB+PtIIYo1BTk9cVWhNPw==","size":{"width":257,"height":24}}$

${"type":"align*","code":"\\begin{align*}\n{I_{2}}&={\\log_{}\\left|x-\\frac{9}{2}+{\\sqrt[]{{\\left(x-\\frac{9}{2}\\right)}^{2}+{\\left(\\frac{1}{2}\\right)}^{2}}}\\right|+C}\\\\\n{I_{2}}&={\\log_{}\\left|x-\\frac{9}{2}+{\\sqrt[]{x^{2}-9x+20}}\\right|+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1601037675867,"cs":"WUGDGVeIog0ahmVjNlUVag==","size":{"width":316,"height":96}}$

Put the value of I1 and I2 in Eq. 2:-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={3.2{\\sqrt[]{x^{2}-9x+20}}+34\\log_{}\\left|x-\\frac{9}{2}+{\\sqrt[]{x^{2}-9x+20}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={6{\\sqrt[]{x^{2}-9x+20}}+34\\log_{}\\left|x-\\frac{9}{2}+{\\sqrt[]{x^{2}-9x+20}}\\right|+C}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-3-1-0","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1601038137183,"cs":"SsICgo2NyEgyeU0rnKup3Q==","size":{"width":480,"height":76}}$

${"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-1-1-0","code":"$20.\\,\\frac{x+2}{{\\sqrt[]{4x-x^{2}}}}$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1600943075081,"cs":"XeV0hpTwutetI7lZv8rvfw==","size":{"width":90,"height":29}}$

Solun:- Let f(x) = ${"type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\frac{x+2}{{\\sqrt[]{4x-x^{2}}}}$","id":"6-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1600943107206,"cs":"xeZtKzqtujDgbOCqMV8+TQ==","size":{"width":48,"height":22}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x+2}{{\\sqrt[]{4x-x^{2}}}}.dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1600943129446,"cs":"VcXXjRUCpriuZVssyrAcEA==","size":{"width":204,"height":37}}$

Using Formula:-

Numerator = A.d(Denominator)/dx + B

${"id":"7-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{x+2}&={A\\diff{\\left(-x^{2}+4x\\right)}{x}+B}\\\\\n{x+2}&={A\\left(-2x+4\\right)+B...\\left(1\\right)}\\\\\n{x+2}&={-2Ax+4A+B}\t\n\\end{align*}","ts":1600943242143,"cs":"YHV4ox7kLG4tnKEpNuEV8w==","size":{"width":210,"height":77}}$

Compare both sides:-

⇒ - 2A = 1 then A = -1/2

⇒ 4A + B = 2

⇒ 4.(-1/2) + B = 2

⇒ B = 4

From eq. 1:-

⇒ x + 2 = (-1/2).(-2x + 4)+4

${"font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\frac{-1}{2}\\left(-2x+4\\right)+4}{{\\sqrt[]{4x-x^{2}}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-1}{2}\\int_{}^{}\\frac{\\left(-2x+4\\right)}{{\\sqrt[]{4x-x^{2}}}}.dx+4\\int_{}^{}\\frac{1}{{\\sqrt[]{4x-x^{2}}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-1}{2}I_{1}+4I_{2}}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-2-1-1-0","ts":1600944497181,"cs":"qX2Yl8PFYEIxyiKzgj4ynQ==","size":{"width":384,"height":126}}$

Calculate I1:-

${"code":"$I_{1}=\\int_{}^{}\\frac{\\left(-2x+4\\right)}{{\\sqrt[]{4x-x^{2}}}}.dx.$","id":"8-0-1-1-0-1-0","font":{"family":"Arial","color":"#222222","size":12},"type":"$","ts":1600944551572,"cs":"cOqBEB65F0mv/a9Oe+135w==","size":{"width":164,"height":32}}$

Let - x2 + 4x = t2

Differentiate w.r.t to t:-

⇒ (-2x + 4).dx = 2t.dt

${"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{I_{1}}&={\\int_{}^{}\\frac{2t.dt}{t}}\\\\\n{I_{1}}&={2\\int_{}^{}1.dt}\t\n\\end{align*}","id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1600944576809,"cs":"qsgxx/nEPKp13WcBBqk4bg==","size":{"width":96,"height":76}}$

We know that:-

${"type":"$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-1-1","code":"$\\int_{}^{}1.dx=x+c$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600941859188,"cs":"WuNgfeVsdBGBqvwBjUS8uw==","size":{"width":100,"height":17}}$

${"font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{I_{1}}&={2t+C}\t\n\\end{align*}","type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-0-1-1-1-0","ts":1600944594159,"cs":"22h2QNIGcm+zaOMnxtEDmA==","size":{"width":80,"height":16}}$

Put the value of t:-

${"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I_{1}}&={2{\\sqrt[]{4x-x^{2}}}+C}\t\n\\end{align*}","type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0","ts":1600944626604,"cs":"oZmImZ6MTQjW5ivRvR/Jbw==","size":{"width":141,"height":21}}$

Calculate I2:-

${"id":"8-1-1-1-1-0","font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{4x-x^{2}}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{-\\left(x^{2}-4x+\\left(2\\right)^{2}-\\left(2\\right)^{2}\\right)}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{-\\left(x^{2}-4x+\\left(2\\right)^{2}-4\\right)}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(2\\right)^{2}-\\left(x-2\\right)^{2}}}}.dx}\t\n\\end{align*}","type":"align*","ts":1600947052029,"cs":"GF91aZK6mTDJl2TOuuEruA==","size":{"width":276,"height":221}}$

We know that:-

${"type":"$","font":{"color":"#000000","family":"Arial","size":10},"id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-2-1-1-1-0","code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{a^{2}-x^{2}}}}.dx=\\sin^{-1}\\left(\\frac{x}{a}\\right)+c$","ts":1600946536340,"cs":"mf5W7Kt2hjraUYBkiHzDug==","size":{"width":196,"height":22}}$

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-1-1-1-1-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I_{2}}&={\\sin^{-1}\\left(\\frac{x-2}{2}\\right)+C}\t\n\\end{align*}","ts":1600947108742,"cs":"ZdKXZ4ryKT1auhy/Lwl++g==","size":{"width":169,"height":37}}$

Put the value of I1 and I2 in Eq. 2:-

${"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-3-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-1}{2}.2{\\sqrt[]{4x-x^{2}}}+4\\sin^{-1}\\left(\\frac{x-2}{2}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={-{\\sqrt[]{4x-x^{2}}}+4\\sin^{-1}\\left(\\frac{x-2}{2}\\right)+C}\t\n\\end{align*}","ts":1601038083074,"cs":"NBW9Ef/OK3H2smB+6GYbhw==","size":{"width":360,"height":80}}$

${"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","code":"$21.\\,\\frac{x+2}{{\\sqrt[]{x^{2}+2x+3}}}$","font":{"color":"#000000","size":12,"family":"Arial"},"ts":1600947272161,"cs":"jZn712SkSiQUr74ZExeMhg==","size":{"width":109,"height":29}}$

Solun:- Let f(x) = ${"code":"$\\frac{x+2}{{\\sqrt[]{x^{2}+2x+3}}}$","type":"$","id":"6-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1600947287990,"cs":"I0zPgwkdo5cG5LDqd9bagg==","size":{"width":61,"height":22}}$

Integrate f(x):-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x+2}{{\\sqrt[]{x^{2}+2x+3}}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1600947309306,"cs":"Uz9kQhlXXmgj70h3LzejXg==","size":{"width":230,"height":37}}$

Using Formula:-

Numerator = A.d(Denominator)/dx + B

${"id":"7-1-1-1-1-0","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{x+2}&={A\\diff{\\left(x^{2}+2x+3\\right)}{x}+B}\\\\\n{x+2}&={A\\left(2x+2\\right)+B...\\left(1\\right)}\\\\\n{x+2}&={2Ax+2A+B}\t\n\\end{align*}","ts":1600947399263,"cs":"60t4l8CtaCQVcPNENUYhyQ==","size":{"width":209,"height":77}}$

Comare both side:-

⇒ 2A = 1 then A = 1/2

⇒ 2A + B = 2

⇒ 2.(1/2) + B = 2

⇒ B = 1

From eq. 1:-

⇒ x + 2 = (1/2).(2x + 2) + 1

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-2-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\frac{1}{2}\\left(2x+2\\right)+1}{{\\sqrt[]{x^{2}+2x+3}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\frac{\\left(2x+2\\right)}{{\\sqrt[]{x^{2}+2x+3}}}.dx+\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+2x+3}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}I_{1}+I_{2}}\t\n\\end{align*}","type":"align*","ts":1600947567210,"cs":"BbSIdUykY7smE60RMk9Hzg==","size":{"width":412,"height":126}}$

Calculate I1:-

${"code":"$I_{1}=\\int_{}^{}\\frac{\\left(2x+2\\right)}{{\\sqrt[]{x^{2}+2x+3}}}.dx.$","font":{"family":"Arial","size":12,"color":"#222222"},"id":"8-0-1-1-0-1-1-0","type":"$","ts":1600947639051,"cs":"zsLnndB7Qa0zq6NzJbNWEA==","size":{"width":180,"height":32}}$

Let x2 + 2x + 3 = t2

Differentiate w.r.t to t:-

⇒ (2x + 2).dx = 2t.dt

${"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{I_{1}}&={\\int_{}^{}\\frac{2t.dt}{t}}\\\\\n{I_{1}}&={2\\int_{}^{}1.dt}\t\n\\end{align*}","id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1600944576809,"cs":"RyKxIrMgQLjlf3jONyP++A==","size":{"width":96,"height":76}}$

We know that:-

${"type":"$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-1-2","code":"$\\int_{}^{}1.dx=x+c$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600941859188,"cs":"WXgKGcewjDOdvY0QxmQ/Wg==","size":{"width":100,"height":17}}$

${"font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{I_{1}}&={2t+C}\t\n\\end{align*}","type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-0-1-1-1-1","ts":1600944594159,"cs":"YCsPqita89nW8mwhQv+8Xw==","size":{"width":80,"height":16}}$

Put the value of t:-

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-1-1-1-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{I_{1}}&={2{\\sqrt[]{x^{2}+2x+3}}+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1600947795751,"cs":"mswqTHs1a8sATX5UY+IO0g==","size":{"width":168,"height":21}}$

Calculate I2:-

${"code":"\\begin{align*}\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+2x+3}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+2x+\\left(1\\right)^{2}-\\left(1\\right)^{2}+3}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+2x+\\left(1\\right)^{2}-2}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x+1\\right)^{2}-\\left({\\sqrt[]{2}}\\right)^{2}}}}.dx}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"id":"8-1-1-1-1-1-0","ts":1600947975543,"cs":"da1tjwJ3rXLvFFvGAXCiFg==","size":{"width":272,"height":212}}$

We know that:-

${"id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-2-1-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}-a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}-a^{2}}}\\right|+c$","type":"$","ts":1600948059793,"cs":"xDVKADSuwMpmWVYawbUiZg==","size":{"width":257,"height":24}}$

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-1-1-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{I_{2}}&={\\log_{}\\left|x+1+{\\sqrt[]{{\\left(x+1\\right)}^{2}-2}}\\right|+C}\\\\\n{I_{2}}&={\\log_{}\\left|x+1+{\\sqrt[]{x^{2}+2x+3}}\\right|+C}\t\n\\end{align*}","ts":1601038314202,"cs":"7V1AF8ozzmjYQ79ILI1R3w==","size":{"width":253,"height":68}}$

Put the value of I1 and I2 in Eq. 2:-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}.2{\\sqrt[]{x^{2}+2x+3}}+\\log_{}\\left|x+1+{\\sqrt[]{x^{2}+2x+3}}\\right|+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={{\\sqrt[]{x^{2}+2x+3}}+\\log_{}\\left|x+1+{\\sqrt[]{x^{2}+2x+3}}\\right|+C}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-3-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601038415201,"cs":"9EsfN397N+RYGR72AcuhQA==","size":{"width":445,"height":76}}$

${"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-0","code":"$22.\\,\\frac{x+3}{x^{2}-2x-5}$","type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"ts":1600949649529,"cs":"1ajKSrrO3QYXaqPU6WU5jA==","size":{"width":96,"height":26}}$

Solun:- Let f(x) = ${"code":"$\\frac{x+3}{x^{2}-2x-5}$","font":{"family":"Arial","color":"#000000","size":10},"id":"6-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","ts":1600949680762,"cs":"B12Sqi2De8MKm2LPEF8bUw==","size":{"width":52,"height":20}}$

Integrate f(x):-

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x+3}{x^{2}-2x-5}.dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1600949703047,"cs":"O3n+U6wkyYG9RN1PN3Q5Bg==","size":{"width":217,"height":36}}$

Using Formula:-

Numerator = A.d(Denominator)/dx + B

${"code":"\\begin{align*}\n{x+3}&={A\\diff{\\left(x^{2}-2x-5\\right)}{x}+B}\\\\\n{x+3}&={A\\left(2x-2\\right)+B...\\left(1\\right)}\\\\\n{x+3}&={2Ax-2A+B}\t\n\\end{align*}","type":"align*","id":"7-1-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1600949904530,"cs":"JF1Bx/rN4zrPDr60IA8n0A==","size":{"width":209,"height":77}}$

Comare both side:-

⇒ 2A = 1 then A = 1/2

⇒ -2A + B = 3

⇒ -2.(1/2) + B = 3

⇒ B = 4

From eq. 1:-

⇒ x + 3 = (1/2).(2x - 2) + 4

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\frac{1}{2}\\left(2x-2\\right)+4}{x^{2}-2x-5}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\frac{\\left(2x-2\\right)}{x^{2}-2x-5}.dx+4\\int_{}^{}\\frac{1}{x^{2}-2x-5}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}I_{1}+4I_{2}}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-2-1-1-1-1-0","ts":1600953427061,"cs":"NiiVg8kKwrOjbnLTv/waqA==","size":{"width":394,"height":122}}$

Calculate I1:-

${"id":"8-0-1-1-0-1-1-1-0","type":"$","code":"$I_{1}=\\int_{}^{}\\frac{\\left(2x-2\\right)}{x^{2}-2x-5}.dx$","font":{"size":12,"color":"#222222","family":"Arial"},"ts":1600951078320,"cs":"43Gv4q+3a6BoCANsxd4/ag==","size":{"width":161,"height":29}}$

Let x2 - 2x - 5 = t

Differentiate w.r.t to t:-

⇒ (2x - 2).dx = dt

${"code":"\\begin{align*}\n{I_{1}}&={\\int_{}^{}\\frac{dt}{t}}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"align*","ts":1600951167691,"cs":"vC1bEbJP3oTSY1hXKYxLHw==","size":{"width":74,"height":36}}$

We know that:-

${"id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-1-3-0","type":"$","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600951220206,"cs":"+fVaZDM3pCGS7koDKCYNdw==","size":{"width":140,"height":18}}$

${"type":"align*","code":"\\begin{align*}\n{I_{1}}&={\\log_{}\\left|t\\right|+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-0-1-1-1-2-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600951255689,"cs":"PytijJuemFbWIHyowbjAvw==","size":{"width":104,"height":16}}$

Put the value of t:-

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{I_{1}}&={\\log_{}\\left|x^{2}-2x-5\\right|+C}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1600951392856,"cs":"3jabRzTgxQOGV/amefIBOQ==","size":{"width":178,"height":20}}$

Calculate I2:-

${"type":"align*","font":{"family":"Arial","color":"#222222","size":10},"id":"8-1-1-1-1-1-1-0","code":"\\begin{align*}\n{I_{2}}&={\\int_{}^{}\\frac{1}{x^{2}-2x-5}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{x^{2}-2x+\\left(1\\right)^{2}-\\left(1\\right)^{2}-5}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{\\left(x-1\\right)^{2}-6}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{\\left(x-1\\right)^{2}-\\left({\\sqrt[]{6}}\\right)^{2}}.dx}\t\n\\end{align*}","ts":1600951702668,"cs":"953a/gfI+UunsAHnRQKNig==","size":{"width":256,"height":181}}$

We know that:-

${"id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","code":"$\\int_{}^{}\\frac{1}{x^{2}-a^{2}}.dx=\\frac{1}{2a}\\log_{}\\left|\\frac{x-a}{x+a}\\right|+c$","ts":1600951920475,"cs":"md4tZ/on4z4wGh2PUPinzA==","size":{"width":202,"height":20}}$

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{I_{2}}&={\\frac{1}{2{\\sqrt[]{6}}}\\log_{}\\left|\\frac{x-1-{\\sqrt[]{6}}}{x-1+{\\sqrt[]{6}}}\\right|+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1600952715091,"cs":"3dGUDu+XlCv0FkB19BiMVA==","size":{"width":220,"height":44}}$

Put the value of I1 and I2 in Eq. 2:-

${"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-3-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left|x^{2}-2x-5\\right|+\\frac{4}{2{\\sqrt[]{6}}}\\log_{}\\left|\\frac{x-1-{\\sqrt[]{6}}}{x-1+{\\sqrt[]{6}}}\\right|}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\log_{}\\left|x^{2}-2x-5\\right|+\\frac{2}{{\\sqrt[]{6}}}\\log_{}\\left|\\frac{x-1-{\\sqrt[]{6}}}{x-1+{\\sqrt[]{6}}}\\right|}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1600953093525,"cs":"5G+1xdV8FTd8RtrlgEpCfA==","size":{"width":389,"height":92}}$

${"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1","code":"$23.\\,\\frac{5x+3}{{\\sqrt[]{x^{2}+4x+10}}}$","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1600953912153,"cs":"RmsYp4aXb6IASaZ+jkWiuw==","size":{"width":116,"height":29}}$

Solun:- Let f(x) = ${"type":"$","id":"6-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\frac{5x+3}{{\\sqrt[]{x^{2}+4x+10}}}$","ts":1600955327151,"cs":"3agA7ywRubijy6R4E46krA==","size":{"width":68,"height":22}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{5x+3}{{\\sqrt[]{x^{2}+4x+10}}}.dx}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601038868809,"cs":"nYDbq+nymNMYebJbfUUJrw==","size":{"width":238,"height":37}}$

Using Formula:-

Numerator = A.d(Denominator)/dx + B

${"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"7-1-1-1-1-1-1","code":"\\begin{align*}\n{5x+3}&={A\\diff{\\left(x^{2}+4x+10\\right)}{x}+B}\\\\\n{5x+3}&={A\\left(2x+4\\right)+B...\\left(1\\right)}\\\\\n{5x+3}&={2Ax+4A+B}\t\n\\end{align*}","ts":1601038940514,"cs":"wWUrPfwdtxO7XVYf5zyoPA==","size":{"width":225,"height":77}}$

Comare both side:-

⇒ 2A = 5 then A = 5/2

⇒ 4A + B = 3

⇒ 4.(5/2) + B = 3

⇒ B = - 7

From eq. 1:-

⇒ 5x + 3 = (5/2).(2x + 4) + (-7)

${"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-2-1-1-1-1-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\frac{5}{2}\\left(2x+4\\right)-7}{{\\sqrt[]{x^{2}+4x+10}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{5}{2}\\int_{}^{}\\frac{\\left(2x+4\\right)}{{\\sqrt[]{x^{2}+4x+10}}}.dx-7\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+4x+10}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{5}{2}I_{1}-7I_{2}}\t\n\\end{align*}","ts":1601039278339,"cs":"DZ3cYD/L7dWuu+YbxabDNw==","size":{"width":437,"height":126}}$

Calculate I1:-

${"code":"$I_{1}=\\int_{}^{}\\frac{\\left(2x+4\\right)}{{\\sqrt[]{x^{2}+4x+10}}}.dx$","id":"8-0-1-1-0-1-1-1-1","type":"$","font":{"family":"Arial","color":"#222222","size":12},"ts":1601039771144,"cs":"OIouELBABshXwyPnt3SKcg==","size":{"width":180,"height":32}}$

Let x2 + 4x + 10 = t2

Differentiate w.r.t to t:-

⇒ (2x + 4).dx = 2t.dt

${"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I_{1}}&={\\int_{}^{}\\frac{2t.dt}{t}}\\\\\n{I_{1}}&={\\int_{}^{}2.dt}\\\\\n{I_{1}}&={2\\int_{}^{}1.dt}\t\n\\end{align*}","id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1","type":"align*","ts":1601040554484,"cs":"4C8yQvkcJ0wyjwNjTMpZ+w==","size":{"width":96,"height":116}}$

We know that:-

${"type":"$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-1-1-1-3-1","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$\\int_{}^{}1.dx=x+c$","ts":1601040577269,"cs":"fI+pyD6+hXa4x3ySfLbXlA==","size":{"width":100,"height":17}}$

${"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-0-1-1-1-2-1","type":"align*","code":"\\begin{align*}\n{I_{1}}&={2t+C}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1601040635580,"cs":"ulrcyZ69w2O71CjM4Ycyeg==","size":{"width":80,"height":16}}$

Put the value of t:-

${"type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-1-1-1-1-1-1-1-1-1","code":"\\begin{align*}\n{I_{1}}&={2{\\sqrt[]{x^{2}+4x+10}}+C}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1601040668063,"cs":"Nkgg1y4UIkt0Ebu8NJjiUQ==","size":{"width":176,"height":21}}$

Calculate I2:-

${"font":{"family":"Arial","color":"#222222","size":10},"type":"align*","id":"8-1-1-1-1-1-1-1","code":"\\begin{align*}\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+4x+10}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+4x+\\left(2\\right)^{2}-\\left(2\\right)^{2}+10}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x+2\\right)^{2}+10}}}.dx}\\\\\n{I_{2}}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(x+2\\right)^{2}+\\left({\\sqrt[]{10}}\\right)^{2}}}}.dx}\t\n\\end{align*}","ts":1601040239765,"cs":"zqOouM+7eg6g207ml84rWQ==","size":{"width":280,"height":212}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{x^{2}+a^{2}}}}.dx=\\log_{}\\left|x+{\\sqrt[]{x^{2}+a^{2}}}\\right|+c$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-1","type":"$","ts":1601040290901,"cs":"oSpVr7HcnRCDiA2MyQQhmw==","size":{"width":257,"height":24}}$

${"code":"\\begin{align*}\n{I_{2}}&={\\log_{}\\left|x+2+{\\sqrt[]{x^{2}+4x+10}}\\right|+C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-1-1-0-1-0-1-0-1-1-1-1-1-1-1-1-1","ts":1601040332235,"cs":"hXjSiBtB1p/TVj/XTQQbMg==","size":{"width":258,"height":26}}$

Put the value of I1 and I2 in Eq. 2:-

${"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-3-1-1-1-1-1","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{5}{2}.2{\\sqrt[]{x^{2}+4x+10}}-7\\log_{}\\left|x+2+{\\sqrt[]{x^{2}+4x+10}}\\right|}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={5{\\sqrt[]{x^{2}+4x+10}}-7\\log_{}\\left|x+2+{\\sqrt[]{x^{2}+4x+10}}\\right|}\t\n\\end{align*}","ts":1601040718033,"cs":"jImca+/NV/ShJqimuYqfDA==","size":{"width":440,"height":76}}$

Choose the correct answer in Exercises 24 and 25.

${"code":"$24.\\,\\int_{}^{}\\frac{dx}{x^{2}+2x+2}$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"1-0-0-1-0","type":"$","ts":1601024198251,"cs":"M7Xd8UlbhkZjF4xk/frJyg==","size":{"width":118,"height":26}}$ equals

(A) x.tan-1(x+1)+C

(B) tan-1(x+1)+C

(C) (x+1).tan-1(x)+C

(D) (x+1).tan-1(x)+C

Solun:- Let f(x) = ${"code":"$\\frac{1}{x^{2}+2x+2}$","id":"6-1-1-1-1-1-1-1-1-0-1-0","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1601024359992,"cs":"jhugs+c0HfCLejCK8BRwWw==","size":{"width":52,"height":20}}$

Integrate f(x):-

${"font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-0-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{x^{2}+2x+2}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{x^{2}+2x+\\left(1\\right)^{2}-\\left(1\\right)^{2}+2}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\left(x+1\\right)^{2}+1}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\left(x+1\\right)^{2}+\\left(1\\right)^{2}}.dx}\t\n\\end{align*}","type":"align*","ts":1601024493393,"cs":"BgEAiwpussnAkLhmjpPdGA==","size":{"width":309,"height":169}}$

Let x + 1 = t

Differentiate w.r.t to t:-

⇒ dx = dt

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-0-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{t^{2}+\\left(1\\right)^{2}}}\t\n\\end{align*}","type":"align*","ts":1601024547552,"cs":"k5fQZl1fNjVDhDf09/I66g==","size":{"width":172,"height":40}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{x^{2}+a^{2}}.dx=\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+c$","font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-0-1","ts":1600772250107,"cs":"/R9kSZ26pOLwX/YogeHJ6A==","size":{"width":204,"height":20}}$

${"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{1}\\tan^{-1}\\left(\\frac{t}{1}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\tan^{-1}\\left(t\\right)+C}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-0-1-0","ts":1601024665425,"cs":"zLKI3JFZfmdy0pRKDzZgsg==","size":{"width":212,"height":77}}$

Put the value of t:-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\tan^{-1}\\left(x+1\\right)+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-1-0-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1601025154192,"cs":"M8IchUrgELFiIUePbifQwQ==","size":{"width":208,"height":36}}$

The correct answer is B.

${"font":{"color":"#000000","family":"Arial","size":12},"id":"1-0-0-1-1","type":"$","code":"$25.\\,\\int_{}^{}\\frac{dx}{{\\sqrt[]{9x-4x^{2}}}}$","ts":1601025398665,"cs":"qTFciIQRBb4b3RubuW5chw==","size":{"width":120,"height":29}}$ equals

Solun:- Let f(x) = ${"font":{"color":"#000000","family":"Arial","size":10},"id":"6-1-1-1-1-1-1-1-1-0-1-1","type":"$","code":"$\\frac{1}{{\\sqrt[]{9x-4x^{2}}}}$","ts":1601025422216,"cs":"KdeBqwFDvpgRs4ZmWT/8Yg==","size":{"width":53,"height":22}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{9x-4x^{2}}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{-4\\left(x^{2}-\\frac{9}{4}x\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{-4\\left(x^{2}-\\frac{9}{4}x+\\left(\\frac{9}{8}\\right)^{2}-\\left(\\frac{9}{8}\\right)^{2}\\right)}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\frac{1}{{\\sqrt[]{-\\left[\\left(x-\\frac{9}{8}\\right)^{2}-\\left(\\frac{9}{8}\\right)^{2}\\right]}}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(\\frac{9}{8}\\right)^{2}-\\left(x-\\frac{9}{8}\\right)^{2}}}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601028444308,"cs":"71A8hXeNJBKGE5wt5+IPpw==","size":{"width":356,"height":276}}$

Let x - 9/8 = t

Differentiate w.r.t to t:-

⇒ dx = dt

${"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-1-0-0-0-0-1-1-1-1-1-1-1-1-0-1-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\frac{dt}{{\\sqrt[]{\\left(\\frac{9}{8}\\right)^{2}-t^{2}}}}}\t\n\\end{align*}","ts":1601028468901,"cs":"nscWnIp02YiZEaLsAaMjBQ==","size":{"width":212,"height":50}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{a^{2}-x^{2}}}}.dx=\\sin^{-1}\\left(\\frac{x}{a}\\right)+c$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"3-0-0-0-1-1-1-1-1-0-1-1-1-1-0-2","type":"$","ts":1601028109646,"cs":"gXtblDLUWLp4/VOMO0+BRQ==","size":{"width":196,"height":22}}$

${"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\sin^{-1}\\left(\\frac{t}{\\frac{9}{8}}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\sin^{-1}\\left(\\frac{8t}{9}\\right)+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-0-1-1-0-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","ts":1601028516528,"cs":"XErBZPDfKHw7ADKO63swCg==","size":{"width":216,"height":89}}$

Put the value of t:-

${"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\sin^{-1}\\left(\\frac{8\\left(x-\\frac{9}{8}\\right)}{9}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\sin^{-1}\\left(\\frac{8x-9}{9}\\right)+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-0-0-1-1-0-1-1-2-0-1-1-1-0-0-1-1-0-1","ts":1601028293778,"cs":"Pza9jMleuFcm/8DTfO/l7A==","size":{"width":266,"height":89}}$

The correct answer is B.

Download PDF of Exercise 7.4

NCERT Maths solutions for 12th

Important Note

Exercise 7.4

Integrate the functions in Exercises 1 to 23.

${"id":"1-0-0-0","type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"code":"$1.\\,\\frac{3x^{2}}{x^{6}+1}$","ts":1600679911258,"cs":"yhhcytWV2Zy+8aue22/oIg==","size":{"width":60,"height":28}}$

${"font":{"size":12,"family":"Arial","color":"#000000"},"code":"$2.\\,\\frac{1}{{\\sqrt[]{1+4x^{2}}}}$","id":"1-1-0","type":"$","ts":1600681152705,"cs":"yH+4L+p5Pz3pKEeeX+74fg==","size":{"width":80,"height":29}}$

${"code":"$3.\\,\\frac{1}{{\\sqrt[]{\\left(2-x\\right)^{2}+1}}}$","type":"$","id":"1-0-1","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1600683112115,"cs":"i0taP9DMKBAGNffHrZTMIQ==","size":{"width":104,"height":40}}$

${"id":"1-1-1-0","code":"$4.\\,\\frac{1}{{\\sqrt[]{9-25x^{2}}}}$","font":{"family":"Arial","size":12,"color":"#000000"},"type":"$","ts":1600684422357,"cs":"xA6eoDw2ZnHe4G0RrL7eaQ==","size":{"width":86,"height":29}}$

${"code":"$5.\\,\\frac{3x}{1+2x^{4}}$","id":"1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1600687377773,"cs":"8xMQN4+Ke0lBn4quorfM9A==","size":{"width":68,"height":26}}$

${"id":"1-1-1-1-1-0","code":"$6.\\,\\frac{x^{2}}{1-x^{6}}$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1600689187469,"cs":"kzOo4d+K4EtwxeWSFxC12A==","size":{"width":60,"height":28}}$

${"code":"$7.\\,\\frac{x-1}{{\\sqrt[]{x^{2}-1}}}$","id":"1-1-1-1-1-1-0","type":"$","font":{"family":"Arial","color":"#000000","size":12},"ts":1600692474140,"cs":"2CwAJZcXfAB9vSR9OoXIHg==","size":{"width":72,"height":29}}$

${"id":"1-1-1-1-1-1-1","code":"$8.\\,\\frac{x^{2}}{{\\sqrt[]{x^{6}+a^{6}}}}$","font":{"color":"#000000","size":12,"family":"Arial"},"type":"$","ts":1600760648537,"cs":"tb5PBNiFXdv5SwZDZ5Qspg==","size":{"width":78,"height":32}}$

${"type":"$","code":"$9.\\,\\frac{\\sec^{2}x}{{\\sqrt[]{\\tan^{2}x+4}}}$","font":{"color":"#000000","family":"Arial","size":12},"id":"1-1-1-1-1-1-2-0","ts":1600761684572,"cs":"+Cd8xeNL3a+11edvQ7D3vw==","size":{"width":98,"height":34}}$

${"id":"1-1-1-1-1-1-2-1-0","type":"$","code":"$10.\\,\\frac{1}{{\\sqrt[]{x^{2}+2x+2}}}$","font":{"color":"#000000","size":12,"family":"Arial"},"ts":1600770812168,"cs":"qJDR7XqOm/StXnT2MwYwYQ==","size":{"width":109,"height":29}}$

${"type":"$","font":{"color":"#000000","family":"Arial","size":12},"id":"1-1-1-1-1-1-2-1-1-0","code":"$11.\\,\\frac{1}{9x^{2}+6x+5}$","ts":1600771733608,"cs":"n3yfxe/6OZKUG+r1BZqIUQ==","size":{"width":104,"height":25}}$

${"type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"code":"$12.\\,\\frac{1}{{\\sqrt[]{7-6x-x^{2}}}}$","id":"1-1-1-1-1-1-2-1-1-1-0","ts":1600772635181,"cs":"jhUJnlz62DNQwGe2MHZkqQ==","size":{"width":109,"height":29}}$

${"type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"code":"$13.\\,\\frac{1}{{\\sqrt[]{\\left(x-1\\right)\\left(x-2\\right)}}}$","id":"1-1-1-1-1-1-2-1-1-1-1-0","ts":1600773370112,"cs":"nff5c3OKlKZ5QxUe1TTy6w==","size":{"width":128,"height":32}}$

${"font":{"family":"Arial","size":12,"color":"#000000"},"id":"1-1-1-1-1-1-2-1-1-1-1-1-0","type":"$","code":"$14.\\,\\frac{1}{{\\sqrt[]{8+3x-x^{2}}}}$","ts":1600774667628,"cs":"HtcXntPvWH0zLwgeP3XhTg==","size":{"width":109,"height":29}}$

${"type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-0","code":"$15.\\,\\frac{1}{{\\sqrt[]{\\left(x-a\\right)\\left(x-b\\right)}}}$","ts":1600848398702,"cs":"lVXbcr/8VM4voEPEzTUarw==","size":{"width":128,"height":32}}$

${"code":"$16.\\,\\frac{4x+1}{{\\sqrt[]{2x^{2}+x-3}}}$","id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-0","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1600863321920,"cs":"lbV4IFAKVzpSTUvMPfNIKw==","size":{"width":109,"height":29}}$

${"font":{"size":12,"family":"Arial","color":"#000000"},"type":"$","code":"$17.\\,\\frac{x+2}{{\\sqrt[]{x^{2}-1}}}$","id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0","ts":1600865328985,"cs":"48qkg8YIs+YBPJDJdyqKkQ==","size":{"width":82,"height":29}}$

${"type":"$","font":{"color":"#000000","family":"Arial","size":12},"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-0","code":"$18.\\,\\frac{5x-2}{1+2x+3x^{2}}$","ts":1600937528982,"cs":"nSFB3fD6Je9nwgMUNoxfOw==","size":{"width":104,"height":26}}$

${"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-1-0","type":"$","font":{"family":"Arial","color":"#000000","size":12},"code":"$19.\\,\\frac{6x+7}{{\\sqrt[]{\\left(x-5\\right)\\left(x-4\\right)}}}$","ts":1600943034494,"cs":"9d3+ZMCOp2qlHqBXIEjKWw==","size":{"width":128,"height":32}}$

${"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-1-1-0","code":"$20.\\,\\frac{x+2}{{\\sqrt[]{4x-x^{2}}}}$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1600943075081,"cs":"XeV0hpTwutetI7lZv8rvfw==","size":{"width":90,"height":29}}$

${"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","code":"$21.\\,\\frac{x+2}{{\\sqrt[]{x^{2}+2x+3}}}$","font":{"color":"#000000","size":12,"family":"Arial"},"ts":1600947272161,"cs":"jZn712SkSiQUr74ZExeMhg==","size":{"width":109,"height":29}}$

${"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-0","code":"$22.\\,\\frac{x+3}{x^{2}-2x-5}$","type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"ts":1600949649529,"cs":"1ajKSrrO3QYXaqPU6WU5jA==","size":{"width":96,"height":26}}$

${"id":"1-1-1-1-1-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1","code":"$23.\\,\\frac{5x+3}{{\\sqrt[]{x^{2}+4x+10}}}$","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1600953912153,"cs":"RmsYp4aXb6IASaZ+jkWiuw==","size":{"width":116,"height":29}}$

Choose the correct answer in Exercises 24 and 25.