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Exercise 7.9

Evaluate the definite integrals in Exercises 1 to 20.

{"font":{"size":10,"color":"#000000","family":"Arial"},"id":"4-0","type":"$","code":"$1.\\,\\int_{-1}^{1}\\left(x+1\\right).dx$","ts":1601965258442,"cs":"QPjvyzoOCjnI14o5F0cihA==","size":{"width":112,"height":21}}

Solun:- Let f(x) = x + 1

Integrate f(x):-

{"font":{"family":"Arial","color":"#000000","size":10},"code":"$I=\\int_{-1}^{1}\\left(x+1\\right).dx$","type":"$","id":"2-0-0-0-0-0-0-0-0-0-0-0-0","ts":1601965363189,"cs":"HMLHHI/dRplJFXAYRIPSNQ==","size":{"width":124,"height":21}}

We know that:-

{"type":"$","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","font":{"color":"#000000","family":"Arial","size":10},"id":"3-0-0-0-0-0-0-0-0-0-0","ts":1601965432018,"cs":"mBE0KAnlinq1/g7ob+51PA==","size":{"width":136,"height":21}}

{"type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={\\left[\\frac{x^{2}}{2}+x\\right]_{-1}^{1}}\\\\\n{I}&={\\left[\\left(\\frac{1}{2}+1\\right)-\\left(\\frac{1}{2}-1\\right)\\right]}\\\\\n{I}&={\\left[\\left(\\frac{3}{2}\\right)-\\left(\\frac{-1}{2}\\right)\\right]}\\\\\n{I}&={2}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0","ts":1601965687150,"cs":"U87i2fFwW15eoZpfMrK1bA==","size":{"width":196,"height":149}}

{"type":"$","id":"4-1-0","code":"$2.\\,\\int_{2}^{3}\\frac{1}{x}.dx$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601965735642,"cs":"GZgjpKy67RoUnnWEfw3lUA==","size":{"width":74,"height":20}}

Solun:- Let f(x) = 1/x

Integrate f(x):-

{"type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-0","code":"$I=\\int_{2}^{3}\\frac{1}{x}.dx$","ts":1601965772168,"cs":"9f1ld5AtHvGQGZhba0OxpA==","size":{"width":85,"height":20}}

We know that:-

{"type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}x+C$","id":"3-0-0-0-0-0-0-0-0-0-1-0","ts":1601965813419,"cs":"fgFUQ6hQPv0BPAtnLxUv+Q==","size":{"width":136,"height":18}}

{"code":"\\begin{align*}\n{I}&={\\left[\\log_{}x\\right]_{2}^{3}}\\\\\n{I}&={\\left[\\left(\\log_{}3\\right)-\\left(\\log_{}2\\right)\\right]}\\\\\n{I}&={\\log_{}\\left(\\frac{3}{2}\\right)}\t\n\\end{align*}","type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601965895575,"cs":"kBvGkpVQLKi0SAnmvhfq1A==","size":{"width":144,"height":82}}

{"id":"4-1-1-0","code":"$3.\\,\\int_{1}^{2}\\left(4x^{3}-5x^{2}+6x+9\\right).dx$","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601969661134,"cs":"VqPmwbyzbGD842JEgRFqlw==","size":{"width":202,"height":20}}

Solun:- Let f(x) = 4x3 - 5x2 + 6x + 9

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-0","type":"$","code":"$I=\\int_{1}^{2}\\left(4x^{3}-5x^{2}+6x+9\\right).dx$","font":{"family":"Arial","color":"#000000","size":10},"ts":1601966025033,"cs":"b7ii12YZpy7YE7Z8D56PYg==","size":{"width":214,"height":20}}

We know that:-

{"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","type":"$","id":"3-0-0-0-0-0-0-0-0-0-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1601966121278,"cs":"OLQSRdCYnGdkX37Rfn81Fw==","size":{"width":136,"height":21}}

{"font":{"family":"Arial","color":"#000000","size":10},"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-0","type":"align*","code":"\\begin{align*}\n{I}&={\\left[4\\frac{x^{4}}{4}-5\\frac{x^{3}}{3}+6\\frac{x^{2}}{2}+9x\\right]_{1}^{2}}\\\\\n{I}&={\\left[x^{4}-\\frac{5x^{3}}{3}+3x^{2}+9x\\right]_{1}^{2}}\t\n\\end{align*}","ts":1601969612635,"cs":"ki9v0v3+j9BmV/GATHjlGQ==","size":{"width":217,"height":89}}

{"type":"align*","code":"\\begin{align*}\n{I}&={\\left[\\left(16-\\frac{40}{3}+12+18\\right)-\\left(1-\\frac{5}{3}+3+9\\right)\\right]}\\\\\n{I}&={\\frac{64}{3}}\t\n\\end{align*}","id":"1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601969636754,"cs":"QobS8ElqwkZtU8eDjJarBA==","size":{"width":336,"height":74}}

{"id":"4-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"code":"$4.\\,\\int_{0}^{\\frac{\\Pi}{4}}\\sin2x.dx$","type":"$","ts":1601969723529,"cs":"HUY1TQmobnAG6aRt7ff/SA==","size":{"width":105,"height":24}}

Solun:- Let f(x) = sin 2x

Integrate f(x):-

{"code":"$I=\\int_{0}^{\\frac{\\Pi}{4}}\\sin2x.dx$","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601969743192,"cs":"4mJ3FGy1a/XeB7idOzzRHA==","size":{"width":116,"height":24}}

We know that:-

{"type":"$","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}\\sin x.dx=-\\cos x+C$","id":"3-0-0-0-0-0-0-0-0-0-1-1-1-0","ts":1601969902211,"cs":"1K0qbW3CIn4agdsFjVetRA==","size":{"width":169,"height":17}}

{"code":"\\begin{align*}\n{I}&={-\\left[\\frac{\\cos 2x}{2}\\right]_{0}^{\\frac{\\Pi}{4}}}\\\\\n{I}&={-\\left[\\left(\\frac{\\cos\\frac{\\Pi}{2}}{2}\\right)-\\left(\\frac{\\cos0}{2}\\right)\\right]_{0}^{\\frac{\\Pi}{4}}}\\\\\n{I}&={-\\left[\\left(\\frac{0}{2}\\right)-\\left(\\frac{1}{2}\\right)\\right]}\\\\\n{I}&={\\frac{1}{2}}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1602155973031,"cs":"9oU1Tz0OeWpeAQUkia+rvg==","size":{"width":224,"height":181}}

{"type":"$","code":"$5.\\,\\int_{0}^{\\frac{\\Pi}{2}}\\cos2x.dx$","font":{"color":"#000000","family":"Arial","size":10},"id":"4-1-1-1-1-0","ts":1601970304138,"cs":"RLAW0qE52OlF6cFPypQm0g==","size":{"width":108,"height":24}}

Solun:- Let f(x) = cos 2x

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-0","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$I=\\int_{0}^{\\frac{\\Pi}{2}}\\cos2x.dx$","ts":1601970341548,"cs":"rs58gWyVLW8LpIDz7SAXSA==","size":{"width":118,"height":24}}

We know that:-

{"font":{"color":"#000000","family":"Arial","size":10},"code":"$\\int_{}^{}\\cos x.dx=\\sin x+C$","type":"$","id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-0","ts":1601970366334,"cs":"LbP3u42TrJzjSZUITCjQxg==","size":{"width":154,"height":17}}

{"code":"\\begin{align*}\n{I}&={\\left[\\frac{\\sin2x}{2}\\right]_{0}^{\\frac{\\Pi}{2}}}\\\\\n{I}&={\\left[\\left(\\frac{\\sin\\Pi}{2}\\right)-\\left(\\frac{\\sin0}{2}\\right)\\right]_{0}^{\\frac{\\Pi}{4}}}\\\\\n{I}&={0}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-1-0","type":"align*","ts":1601970528050,"cs":"+PCyTeN6leni/Hzbp7xDpg==","size":{"width":200,"height":113}}

{"code":"$6.\\,\\int_{4}^{5}e^{x}.dx$","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"id":"4-1-1-1-1-1-0","ts":1601970676802,"cs":"eIuz8oq0YmiMPo9bF864xw==","size":{"width":76,"height":20}}

Solun:- Let f(x) = ex

Integrate f(x):-

{"type":"$","font":{"family":"Arial","color":"#000000","size":10},"code":"$I=\\int_{4}^{5}e^{x}.dx$","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-0","ts":1601970697530,"cs":"3IXbZzW5do6Bpv370Zg0bg==","size":{"width":88,"height":20}}

We know that:-

{"id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-0","code":"$\\int_{}^{}e^{x}.dx=e^{x}+C$","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","ts":1601970725585,"cs":"E+bFAu9+z6qM0n3vbCWyiA==","size":{"width":120,"height":17}}

{"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-1-1-0","code":"\\begin{align*}\n{I}&={\\left[e^{x}\\right]_{4}^{5}}\\\\\n{I}&={\\left[\\left(e^{5}\\right)-\\left(e^{4}\\right)\\right]}\\\\\n{I}&={e^{4}\\left(e-1\\right)}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1601970818109,"cs":"5BIeM8Wz1RnCrpDOr+6bYA==","size":{"width":117,"height":66}}

{"font":{"family":"Arial","color":"#000000","size":10},"type":"$","code":"$7.\\,\\int_{0}^{\\frac{\\Pi}{4}}\\tan x.dx$","id":"4-1-1-1-1-1-1-0","ts":1601970936054,"cs":"/VMHEf4cz0fX+Aes8pTwAQ==","size":{"width":100,"height":24}}

Solun:- Let f(x) = tan x

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-0","code":"$I=\\int_{0}^{\\frac{\\Pi}{4}}\\tan x.dx$","type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601970958146,"cs":"HpDOE15e4dLHbtmsgfoaZQ==","size":{"width":112,"height":24}}

We know that:-

{"font":{"family":"Arial","color":"#000000","size":10},"id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-0","code":"$\\int_{}^{}\\tan x.dx=-\\log_{}\\left|\\cos x\\right|+C$","type":"$","ts":1601971359273,"cs":"T0ghDzOduGyCl5IdzZqRGQ==","size":{"width":205,"height":17}}

{"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-1-1-1-0","code":"\\begin{align*}\n{I}&={-\\left[\\log_{}\\left|\\cos x\\right|\\right]_{0}^{\\frac{\\Pi}{4}}}\\\\\n{I}&={-\\left[\\left(\\log_{}\\left|\\cos\\frac{\\Pi}{4}\\right|\\right)-\\left(\\log_{}\\left|\\cos0\\right|\\right)\\right]}\\\\\n{I}&={-\\left[\\left(\\log_{}\\left|\\frac{1}{{\\sqrt[]{2}}}\\right|\\right)-\\left(\\log_{}\\left|1\\right|\\right)\\right]}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1601975834668,"cs":"RHJ6mtG8r6LgnMqOSP0kxQ==","size":{"width":253,"height":112}}

We know that:-

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

log mn = nlog m

{"id":"3","type":"align*","code":"\\begin{align*}\n{I}&={-\\left(\\log_{}\\left(1\\right)-\\log_{}\\left({\\sqrt[]{2}}\\right)\\right)}\\\\\n{I}&={-\\left(0-\\log_{}\\left({\\sqrt[]{2}}\\right)\\right)}\\\\\n{I}&={\\log_{}\\left({\\sqrt[]{2}}\\right)}\\\\\n{I}&={\\frac{1}{2}\\log_{}\\left(2\\right)}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601975916033,"cs":"7Ptc3mkNN6FepwVFhsh84w==","size":{"width":188,"height":132}}

{"font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$8.\\,\\int_{\\frac{\\Pi}{6}}^{\\frac{\\Pi}{4}}\\cos ec\\,\\,x.dx$","id":"4-1-1-1-1-1-1-1-0","ts":1601976013845,"cs":"1i9WSOnnyEKqtghbrjXSzw==","size":{"width":118,"height":29}}

Solun:- Let f(x) = cosec x

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-0","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$I=\\int_{\\frac{\\Pi}{6}}^{\\frac{\\Pi}{4}}\\cos ec\\,\\,x.dx$","ts":1601976055383,"cs":"OY7pWBExSwoXCyzdN+GIxg==","size":{"width":130,"height":29}}

We know that:-

{"id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-0","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$\\int_{}^{}\\cos ec\\,x.dx=\\log_{}\\left|\\cos ec\\,x-\\cot x\\right|+C$","ts":1601976159931,"cs":"H3KmTwWGRsftpHQpB43YEw==","size":{"width":274,"height":17}}

{"type":"align*","code":"\\begin{align*}\n{I}&={\\left[\\log_{}\\left|\\cos ec\\,x-\\cot x\\right|\\right]_{\\frac{\\Pi}{6}}^{\\frac{\\Pi}{4}}}\\\\\n{I}&={\\left[\\left(\\log_{}\\left|\\cos ec\\frac{\\Pi}{4}-\\cot\\frac{\\Pi}{4}\\right|\\right)-\\left(\\log_{}\\left|\\cos ec\\frac{\\Pi}{6}-\\cot\\frac{\\Pi}{6}\\right|\\right)\\right]}\\\\\n{I}&={\\left(\\log_{}\\left|{\\sqrt[]{2}}-1\\right|\\right)-\\left(\\log_{}\\left|2-{\\sqrt[]{3}}\\right|\\right)}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-1-1-1-1-0","ts":1601976405402,"cs":"+zypB1H0o1tt4UxAjNz1AQ==","size":{"width":417,"height":105}}

We know that:-

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

{"type":"align*","id":"4","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={\\log_{}\\left(\\frac{{\\sqrt[]{2}}-1}{2-{\\sqrt[]{3}}}\\right)}\t\n\\end{align*}","ts":1601977472387,"cs":"dgjZYKmdtF0S2sdtJQcOiA==","size":{"width":133,"height":46}}

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","code":"$9.\\,\\int_{0}^{1}\\frac{dx}{{\\sqrt[]{1-x^{2}}}}$","id":"4-1-1-1-1-1-1-1-1-0","ts":1601977718972,"cs":"5cTi+anIqxLo7cV9OoO0fg==","size":{"width":78,"height":24}}

Solun:- Let f(x) = {"type":"$","code":"$\\frac{1}{{\\sqrt[]{1-x^{2}}}}$","id":"5-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1601977786984,"cs":"tyrbDSwXiDBmJisRkpHjnA==","size":{"width":41,"height":22}}

Integrate f(x):-

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-0","type":"$","code":"$I=\\int_{0}^{1}\\frac{dx}{{\\sqrt[]{1-x^{2}}}}$","ts":1601977806544,"cs":"gdNtWGw3eaQEuRohI7MBBw==","size":{"width":89,"height":24}}

We know that:-

{"id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{1-x^{2}}}}.dx=\\sin^{-1}x+C$","ts":1601977891458,"cs":"FzJDPkcgYxavtMWxuhhhmQ==","size":{"width":178,"height":22}}

{"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-1-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{I}&={\\left[\\sin^{-1}x\\right]_{0}^{1}}\\\\\n{I}&={\\left[\\left(\\sin^{-1}\\left(1\\right)\\right)-\\left(\\sin^{-1}\\left(0\\right)\\right)\\right]}\\\\\n{I}&={\\frac{\\Pi}{2}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1601978632196,"cs":"ASCC+rbFKtKTsJ/QZn8RPw==","size":{"width":205,"height":84}}

{"code":"$10.\\,\\int_{0}^{1}\\frac{dx}{1+x^{2}}$","id":"4-1-1-1-1-1-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","ts":1601979050162,"cs":"IZYkCaAgQWOAylWOSZIq3g==","size":{"width":76,"height":22}}

Solun:- Let f(x) = {"code":"$\\frac{1}{1+x^{2}}$","id":"5-1-0","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601979073969,"cs":"AsiXfVU4dkGJS2sbBhVAiw==","size":{"width":32,"height":20}}

Integrate f(x):-

{"code":"$I=\\int_{0}^{1}\\frac{dx}{1+x^{2}}$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-0","ts":1601979089613,"cs":"gK4GxSvgAxcfkBa9IxHBGQ==","size":{"width":80,"height":22}}

We know that:-

{"id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\int_{}^{}\\frac{1}{1+x^{2}}.dx=\\tan^{-1}x+C$","type":"$","ts":1601979238717,"cs":"qeATqvgh1aMt0OKxNEc8/Q==","size":{"width":172,"height":20}}

{"code":"\\begin{align*}\n{I}&={\\left[\\tan^{-1}x\\right]_{0}^{1}}\\\\\n{I}&={\\left[\\left(\\tan^{-1}\\left(1\\right)\\right)-\\left(\\tan^{-1}\\left(0\\right)\\right)\\right]}\\\\\n{I}&={\\frac{\\Pi}{4}}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-1-1-1-1-1-1-0","type":"align*","ts":1601979273752,"cs":"M+WzVDmiC1QRGjgdwIrFhw==","size":{"width":212,"height":84}}

{"id":"4-1-1-1-1-1-1-1-1-1-1-0","type":"$","code":"$11.\\,\\int_{2}^{3}\\frac{dx}{x^{2}-1}$","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601979354042,"cs":"4dAbaaFVzlqnYHnSv0tsMQ==","size":{"width":76,"height":22}}

Solun:- Let f(x) = {"font":{"family":"Arial","color":"#000000","size":10},"type":"$","code":"$\\frac{1}{x^{2}-1}$","id":"5-1-1-0","ts":1601979381508,"cs":"GAGDJK+rsMxebu67uWr08Q==","size":{"width":32,"height":20}}

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-0","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$I=\\int_{2}^{3}\\frac{dx}{x^{2}-1}$","ts":1601979471165,"cs":"YS52Fk0KemExlRjIF35i8w==","size":{"width":80,"height":22}}

By partial fraction:-

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","id":"6","code":"$\\frac{1}{x^{2}-1}=\\frac{A}{\\left(x-1\\right)}+\\frac{B}{\\left(x+1\\right)}$","ts":1601980124960,"cs":"L3UGv9bChZyM36oISo+WzA==","size":{"width":144,"height":22}}

1 = A(x + 1) + B(x - 1).......(1)

Put x = -1 in eq. 1:-

1 = B(- 2)

B = -1/2

Put x = 1 in eq. 1:-

1 = A(2)

A = 1/2

{"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{2}^{3}\\left[\\frac{1}{2}\\frac{1}{\\left(x-1\\right)}-\\frac{1}{2}\\frac{1}{\\left(x+1\\right)}\\right].dx}\\\\\n{I}&={\\frac{1}{2}\\int_{2}^{3}\\left[\\frac{1}{\\left(x-1\\right)}-\\frac{1}{\\left(x+1\\right)}\\right].dx}\\\\\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1601981353077,"cs":"XmrC1L1LNl8v3jeevDn6ZA==","size":{"width":252,"height":84}}

We know that:-

{"id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+C$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","ts":1601980607602,"cs":"VK6/ts6g/BQZ3sEYOOhovA==","size":{"width":145,"height":18}}

{"type":"align*","code":"\\begin{align*}\n{I}&={\\frac{1}{2}\\left[\\log_{}\\left|x-1\\right|-\\log_{}\\left|x+1\\right|\\right]_{2}^{3}}\\\\\n{I}&={\\frac{1}{2}\\left[\\left(\\log_{}\\left|2\\right|-\\log_{}\\left|4\\right|\\right)-\\left(\\log_{}\\left|1\\right|-\\log_{}\\left|3\\right|\\right)\\right]}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-1-1-1-1-1-1-1-0-0","ts":1601981441152,"cs":"Ag9s3BDpSi45a2m41Y5Fng==","size":{"width":296,"height":68}}

We know that:-

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

{"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-1-1-1-1-1-1-1-1","code":"\\begin{align*}\n{I}&={\\frac{1}{2}\\left[\\left(\\log_{}\\left|\\frac{2}{4}\\right|\\right)-\\left(\\log_{}\\left|\\frac{1}{3}\\right|\\right)\\right]}\\\\\n{I}&={\\frac{1}{2}\\left[\\log_{}\\left|\\frac{2}{4}\\right|-\\log_{}\\left|\\frac{1}{3}\\right|\\right]}\\\\\n{I}&={\\frac{1}{2}\\log_{}\\left|\\frac{\\frac{2}{4}}{\\frac{1}{3}}\\right|=\\frac{1}{2}\\log_{}\\left|\\frac{\\frac{1}{2}}{\\frac{1}{3}}\\right|}\\\\\n{I}&={\\frac{1}{2}\\log_{}\\left|\\frac{3}{2}\\right|}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1601981639058,"cs":"APqW7TMR2MmbALh7P2PjVg==","size":{"width":224,"height":169}}

{"code":"$12.\\,\\int_{0}^{\\frac{\\Pi}{2}}\\cos^{2}x.dx$","type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"4-1-1-1-1-1-1-1-1-1-1-1-0","ts":1601981833242,"cs":"c4h3wTl1/OYAT2Bmzt/aIg==","size":{"width":113,"height":24}}

Solun:- Let f(x) = cos2x

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-0","type":"$","code":"$I=\\int_{0}^{\\frac{\\Pi}{2}}\\cos^{2}x.dx$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601981867188,"cs":"Z7rDqKNKOYDJx5K5ebybtA==","size":{"width":117,"height":24}}

We know that:-

cos 2x = 2cos2x - 1

{"code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}\\frac{1+\\cos2x}{2}.dx}\\\\\n{I}&={\\frac{1}{2}\\int_{0}^{\\frac{\\Pi}{2}}\\left(1+\\cos2x\\right).dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601982149396,"cs":"/9Mc3YfkKfJCGdQy3tVWuA==","size":{"width":178,"height":88}}

We know that:-

{"id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\int_{}^{}\\cos x.dx=\\sin x+C$","ts":1601982176270,"cs":"Sui8ddYWgSNKA3VOatQuEw==","size":{"width":154,"height":17}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-0","type":"align*","code":"\\begin{align*}\n{I}&={\\frac{1}{2}\\left[x+\\frac{\\sin2x}{2}\\right]_{0}^{\\frac{\\Pi}{2}}}\\\\\n{I}&={\\frac{1}{2}\\left[\\left(\\frac{\\Pi}{2}+\\frac{1}{2}\\sin\\Pi\\right)-\\left(0+\\frac{1}{2}\\sin0\\right)\\right]}\\\\\n{I}&={\\frac{\\Pi}{4}}\t\n\\end{align*}","ts":1601982344066,"cs":"myUYTMnlW+Hc8aRB6QIMXQ==","size":{"width":285,"height":124}}

{"type":"$","code":"$13.\\,\\int_{2}^{3}\\frac{x}{x^{2}+1}.dx$","font":{"family":"Arial","color":"#000000","size":10},"id":"4-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1601982516907,"cs":"hftcn4vMuvsR1FjUK/jdSA==","size":{"width":101,"height":22}}

Solun:- Let f(x) = {"code":"$\\frac{x}{x^{2}+1}$","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"7-0","ts":1601982580510,"cs":"fZZnk3VSf4a9m3AN2Wa9mQ==","size":{"width":32,"height":17}}

Integrate f(x):-

{"code":"$I=\\int_{2}^{3}\\frac{x}{x^{2}+1}.dx$","type":"$","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-0","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601982686116,"cs":"9ekAJ91tzwugxx7oOmi6Mw==","size":{"width":104,"height":22}}

Let x2 + 1 = t……(1)

⇒ 2x.dx = dt

⇒ x.dx = (1/2).dt

Change limits:

Put x = 2 in eq. 1:-

⇒ t = 5

Put x = 3 in eq. 1:-

⇒ t = 10

{"code":"\\begin{align*}\n{I}&={\\int_{5}^{10}\\frac{1}{2}.\\frac{1}{t}.dt}\\\\\n{I}&={\\frac{1}{2}\\int_{5}^{10}\\frac{1}{t}.dt}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-0","ts":1601983361696,"cs":"dQNP+aGnxdMCJz5TjZmSOA==","size":{"width":117,"height":82}}

We know that:-

{"id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+C$","ts":1601983394500,"cs":"jiHy88Xx5R1mNr0bZQRz+w==","size":{"width":145,"height":18}}

{"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-0","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{I}&={\\frac{1}{2}\\left[\\log_{}\\left|t\\right|\\right]_{5}^{10}}\\\\\n{I}&={\\frac{1}{2}\\left[\\left(\\log_{}\\left|10\\right|\\right)-\\left(\\log_{}\\left|5\\right|\\right)\\right]}\t\n\\end{align*}","type":"align*","ts":1601983453629,"cs":"1ZOWXQlbYKT+J9XmXSTHkg==","size":{"width":185,"height":68}}

We know that:-

⇒ log m - log n = log m/n

{"id":"3-0-0-0-0-0-0-1-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1","type":"align*","code":"\\begin{align*}\n{I}&={\\frac{1}{2}\\log_{}\\left|\\frac{10}{5}\\right|}\\\\\n{I}&={\\frac{1}{2}\\log_{}\\left|2\\right|}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1601983567200,"cs":"fL+9JUEKgCZpKhJpyR4lnQ==","size":{"width":100,"height":70}}

{"code":"$14.\\,\\int_{0}^{1}\\frac{2x+3}{5x^{2}+1}.dx$","type":"$","font":{"color":"#000000","family":"Arial","size":10},"id":"4-1-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1601983766984,"cs":"cKg5XTzu/bsyo63OvysOuQ==","size":{"width":106,"height":22}}

Solun:- Let f(x) = {"id":"7-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$\\frac{2x+3}{5x^{2}+1}$","ts":1601983809954,"cs":"DgQIXaBp1j60H977b6yXmA==","size":{"width":37,"height":20}}

Integrate f(x):-

{"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{0}^{1}\\frac{2x+3}{5x^{2}+1}.dx}\\\\\n{I}&={2\\int_{0}^{1}\\frac{x}{5x^{2}+1}.dx+\\int_{0}^{1}\\frac{3}{5x^{2}+1}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1601983993724,"cs":"fjguDIiCQZAKx7iuDjpSEw==","size":{"width":272,"height":82}}

Let 5x2 + 1 = t……(1)

⇒ 10x.dx = dt

⇒ x.dx = (1/10).dt

Change limits:

Put x = 0 in eq. 1:-

⇒ t = 1

Put x = 1 in eq. 1:-

⇒ t = 6

{"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{I}&={2\\int_{1}^{6}\\frac{1}{t}.\\frac{dt}{10}+\\int_{0}^{1}\\frac{3}{5x^{2}+1}.dx}\\\\\n{I}&={\\frac{1}{5}\\int_{1}^{6}\\frac{1}{t}.dt+\\frac{3}{5}\\int_{0}^{1}\\frac{1}{x^{2}+\\frac{1}{5}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-0","ts":1601984386427,"cs":"2aWY+J3xAn1PE3PGAvAPdg==","size":{"width":248,"height":86}}

We know that:-

{"id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+C$","ts":1601983394500,"cs":"iRrI145pM6isSof7foHKqg==","size":{"width":145,"height":18}}

{"code":"$\\int_{}^{}\\frac{1}{x^{2}+a^{2}}.dx=\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+C$","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"8-0","ts":1601985873970,"cs":"QA3p47mXnFB27ASkNCCYWQ==","size":{"width":208,"height":20}}

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{I}&={\\frac{1}{5}\\left[\\log_{}\\left|t\\right|\\right]_{1}^{6}+\\frac{3}{5}.\\left[\\frac{1}{\\frac{1}{{\\sqrt[]{5}}}}\\tan^{-1}\\left(\\frac{x}{\\frac{1}{{\\sqrt[]{5}}}}\\right)\\right]_{0}^{1}}\\\\\n{I}&={\\frac{1}{5}\\left[\\log_{}\\left|t\\right|\\right]_{1}^{6}+\\frac{3}{5}.\\left[{\\sqrt[]{5}}\\tan^{-1}\\left(x{\\sqrt[]{5}}\\right)\\right]_{0}^{1}}\\\\\n{I}&={\\frac{1}{5}\\left[\\log_{}\\left|6\\right|-\\log_{}\\left|1\\right|\\right]+\\frac{3{\\sqrt[]{5}}}{5}\\left[\\left(\\tan^{-1}\\left({\\sqrt[]{5}}\\right)\\right)-\\left(\\tan^{-1}\\left(0\\times{\\sqrt[]{5}}\\right)\\right)\\right]}\\\\\n{I}&={\\frac{1}{5}\\log_{}\\left|6\\right|+\\frac{3}{{\\sqrt[]{5}}}\\tan^{-1}\\left({\\sqrt[]{5}}\\right)}\t\n\\end{align*}","font":{"size":9,"family":"Arial","color":"#000000"},"ts":1601987184042,"cs":"vr1xlHGAgpoof+xZDnrfFg==","size":{"width":469,"height":177}}

{"id":"4-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","code":"$15.\\,\\int_{0}^{1}x.e^{x^{2}}.dx$","font":{"family":"Arial","color":"#000000","size":10},"ts":1601987473442,"cs":"SlRUqPDEgDwdSyZEvyRV7w==","size":{"width":104,"height":21}}

Solun:- Let f(x) = {"code":"$x.e^{x^{2}}$","type":"$","id":"7-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601987576065,"cs":"vQVTA+XrcuMZNQIcNyRIEw==","size":{"width":34,"height":14}}

Integrate f(x):-

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{I}&={\\int_{0}^{1}x.e^{x^{2}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-0","ts":1601987608896,"cs":"pGV3l+Dlt9MHJA7T5S1RTw==","size":{"width":113,"height":38}}

Let x2 = t……(1)

⇒ 2x.dx = dt

⇒ x.dx = (1/2).dt

Change limits:

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = 1 in eq. 1:-

⇒ t = 1

{"code":"\\begin{align*}\n{I}&={\\int_{0}^{1}e^{t}.\\frac{dt}{2}}\\\\\n{I}&={\\frac{1}{2}\\int_{0}^{1}e^{t}.dt}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1601987943400,"cs":"wrRCTqlQn+LWNb8vlIEMfw==","size":{"width":104,"height":82}}

We know that:-

{"id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\int_{}^{}e^{x}.dx=e^{x}+C$","type":"$","ts":1601987974731,"cs":"EEhM+0ewobrupzJ10YVYmQ==","size":{"width":120,"height":17}}

{"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1-1","code":"\\begin{align*}\n{I}&={\\frac{1}{2}\\left[e^{t}\\right]_{0}^{1}}\\\\\n{I}&={\\frac{1}{2}\\left(e-e^{0}\\right)}\\\\\n{I}&={\\frac{1}{2}\\left(e-1\\right)}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601988331094,"cs":"UTybBJf1Ws/+Gtqo0Mo/gw==","size":{"width":98,"height":105}}

{"code":"$16.\\,\\int_{1}^{2}\\frac{5x^{2}}{x^{2}+4x+3}.dx$","font":{"family":"Arial","color":"#000000","size":10},"id":"4-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","ts":1601989550607,"cs":"sMcHJDQYAwbPDp0uA+/kZA==","size":{"width":121,"height":22}}

Solun:- Let f(x) = {"type":"$","id":"7-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\frac{5x^{2}}{x^{2}+4x+3}$","ts":1601989203020,"cs":"YhGMvNcdU/xKykDHYmz2Wg==","size":{"width":52,"height":21}}

Integrate f(x):-

{"code":"\\begin{align*}\n{I}&={\\int_{1}^{2}\\frac{5x^{2}}{x^{2}+4x+3}.dx}\\\\\n{I}&={\\int_{1}^{2}5.dx-\\int_{1}^{2}\\frac{20x+15}{x^{2}+4x+3}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-1-0-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1601989492580,"cs":"jQTxvWYE1mUGNJ4Ae//Wmg==","size":{"width":238,"height":82}}

We know that:-

⇒ Numerator = A[d(Denominator)/dx]+B

{"code":"\\begin{align*}\n{20x+15}&={A\\diff{\\left(x^{2}+4x+3\\right)}{x}+B}\\\\\n{20x+15}&={A\\left(2x+4\\right)+B....\\left(1\\right)}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"9","ts":1602059449226,"cs":"sAQd0DJc4mdNBu2yBV8N+w==","size":{"width":233,"height":57}}

Compare both sides:-

⇒ 2A = 20 then A = 10

⇒ 4A+B = 15

⇒ 4(10)+B = 15

⇒ B = - 25

From eq. 1:-

Then (20x + 15) = 10(2x+4) - 25

{"code":"\\begin{align*}\n{I}&={\\int_{1}^{2}5.dx-\\left[\\int_{1}^{2}\\frac{10\\left(2x+4\\right)}{x^{2}+4x+3}.dx-\\int_{1}^{2}\\frac{25}{x^{2}+4x+3}.dx\\right]}\\\\\n{I}&={\\int_{1}^{2}5.dx-\\int_{1}^{2}\\frac{10\\left(2x+4\\right)}{x^{2}+4x+3}.dx+\\int_{1}^{2}\\frac{25}{x^{2}+4x+3}.dx}\\\\\n{I}&={\\int_{1}^{2}5.dx-I_{1}+\\int_{1}^{2}\\frac{25}{x^{2}+4x+3}.dx....\\left(2\\right)}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-1-1-0","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1602063576475,"cs":"4c4eTRYYvpid66GSIpzSnA==","size":{"width":409,"height":126}}

Calculating I1:-

{"type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"code":"\\begin{align*}\n{I_{1}}&={\\int_{1}^{2}\\frac{10\\left(2x+4\\right)}{x^{2}+4x+3}.dx}\\\\\n{I_{1}}&={10\\int_{1}^{2}\\frac{2x+4}{x^{2}+4x+3}.dx}\t\n\\end{align*}","id":"10","ts":1602061349607,"cs":"BKk83NFIVvUl9tE3a4DbsQ==","size":{"width":188,"height":82}}

Let x2+4x+3 = t……(3)

⇒ (2x+4).dx = dt

Change limits:

Put x = 1 in eq. 3:-

⇒ t = 8

Put x = 2 in eq. 3:-

⇒ t = 15

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1-2","code":"\\begin{align*}\n{I_{1}}&={10\\int_{8}^{15}\\frac{dt}{t}}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1602072921190,"cs":"47IZanS8QwLtqCO5TIHIXA==","size":{"width":104,"height":38}}

We know that:-

{"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+C$","id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1602061982583,"cs":"kfAMVzL5zojsmGNG3hjcpQ==","size":{"width":145,"height":18}}

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1-3-0","code":"\\begin{align*}\n{I_{1}}&={10\\left[\\log_{}\\left|t\\right|\\right]_{8}^{15}}\\\\\n{I_{1}}&={10\\left(\\log_{}\\left|15\\right|-\\log_{}\\left|8\\right|\\right)}\\\\\n{I_{1}}&={10\\log_{}\\left|\\frac{15}{8}\\right|}\t\n\\end{align*}","ts":1602160946400,"cs":"9xwadc2a/NJTrW235U8zvg==","size":{"width":170,"height":80}}

Put the value of I1 in eq. 2:-

{"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{1}^{2}5.dx-10\\log_{}\\left|\\frac{15}{8}\\right|+25\\int_{1}^{2}\\frac{1}{x^{2}+4x+3}.dx}\\\\\n{I}&={\\int_{1}^{2}5.dx-10\\log_{}\\left|\\frac{15}{8}\\right|+25\\int_{1}^{2}\\frac{1}{x^{2}+4x+\\left(2\\right)^{2}-\\left(2\\right)^{2}+3}.dx}\\\\\n{I}&={\\int_{1}^{2}5.dx-10\\log_{}\\left|\\frac{15}{8}\\right|+25\\int_{1}^{2}\\frac{1}{\\left(x+2\\right)^{2}-1}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1602073143565,"cs":"lafG1vVLgmm9VtZkbXJscA==","size":{"width":444,"height":133}}

We know that:-

{"type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\int_{}^{}\\frac{1}{x^{2}-a^{2}}.dx=\\frac{1}{2a}\\log_{}\\left|\\frac{x-a}{x+a}\\right|$","id":"11","ts":1602064497370,"cs":"9amQeDl24iXCHs8exwsXFA==","size":{"width":176,"height":20}}

{"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-1-1-1-1","code":"\\begin{align*}\n{I}&={\\left[5x\\right]_{1}^{2}-10\\log_{}\\left|\\frac{15}{8}\\right|+\\frac{25}{2}\\left[\\log_{}\\left|\\frac{x+2-1}{x+2+1}\\right|\\right]_{1}^{2}}\\\\\n{I}&={\\left[5x\\right]_{1}^{2}-10\\log_{}\\left|\\frac{15}{8}\\right|+\\frac{25}{2}\\left[\\log_{}\\left|\\frac{x+1}{x+3}\\right|\\right]_{1}^{2}}\\\\\n{I}&={\\left[10-5\\right]-10\\log_{}\\left|\\frac{15}{8}\\right|+\\frac{25}{2}\\left[\\left(\\log_{}\\left|\\frac{2+1}{2+3}\\right|\\right)-\\left(\\log_{}\\left|\\frac{1+1}{1+3}\\right|\\right)\\right]}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1602073292854,"cs":"ISulqX0fH8RqAsNr9qCMkw==","size":{"width":453,"height":130}}

{"id":"12-0-0","code":"\\begin{align*}\n{I}&={5-10\\log_{}\\left|\\frac{15}{8}\\right|+\\frac{25}{2}\\left[\\left(\\log_{}\\left|\\frac{3}{5}\\right|\\right)-\\left(\\log_{}\\left|\\frac{2}{4}\\right|\\right)\\right]}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1602073320938,"cs":"9M36uimnHzzch+Kws+Rhrw==","size":{"width":354,"height":37}}

We know that:-

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

{"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{I}&={5-10\\left(\\log_{}\\left|15\\right|-\\log_{}\\left|8\\right|\\right)+\\frac{25}{2}\\left[\\left(\\log_{}\\left|3\\right|-\\log_{}\\left|5\\right|\\right)-\\left(\\log_{}\\left|2\\right|-\\log_{}\\left|4\\right|\\right)\\right]}\\\\\n{I}&={5-10\\log_{}\\left|15\\right|+10\\log_{}\\left|8\\right|+\\frac{25}{2}\\log_{}\\left|3\\right|-\\frac{25}{2}\\log_{}\\left|5\\right|-\\frac{25}{2}\\log_{}\\left|2\\right|+\\frac{25}{2}\\log_{}\\left|4\\right|}\\\\\n{I}&={5-10\\log_{}\\left|5\\times3\\right|+10\\log_{}\\left|4\\times2\\right|+\\frac{25}{2}\\log_{}\\left|3\\right|-\\frac{25}{2}\\log_{}\\left|5\\right|-\\frac{25}{2}\\log_{}\\left|2\\right|+\\frac{25}{2}\\log_{}\\left|4\\right|}\\\\\n{I}&={5-10\\log_{}\\left|5\\right|-10\\log_{}\\left|3\\right|+10\\log_{}\\left|4\\right|+10\\log_{}\\left|2\\right|+\\frac{25}{2}\\log_{}\\left|3\\right|-\\frac{25}{2}\\log_{}\\left|5\\right|-\\frac{25}{2}\\log_{}\\left|2\\right|+\\frac{25}{2}\\log_{}\\left|4\\right|}\\\\\n{I}&={5-\\frac{45}{2}\\log_{}\\left|5\\right|+\\frac{45}{2}\\log_{}\\left|4\\right|+\\frac{5}{2}\\log_{}\\left|3\\right|-\\frac{5}{2}\\log_{}\\left|2\\right|}\\\\\n{I}&={5-\\frac{45}{2}\\left(\\log_{}\\left|5\\right|-\\log_{}\\left|4\\right|\\right)+\\frac{5}{2}\\left(\\log_{}\\left|3\\right|-\\log_{}\\left|2\\right|\\right)}\\\\\n{I}&={5-\\frac{45}{2}\\log_{}\\left|\\frac{5}{4}\\right|+\\frac{5}{2}\\log_{}\\left|\\frac{3}{2}\\right|}\\\\\n{I}&={5-\\frac{5}{2}\\left(9\\log_{}\\left|\\frac{5}{4}\\right|-\\log_{}\\left|\\frac{3}{2}\\right|\\right)}\t\n\\end{align*}","id":"12-0-1","ts":1602162148439,"cs":"wiGqMw4FB3K2JqTBiEYyZw==","size":{"width":680,"height":298}}

{"code":"$17.\\,\\int_{0}^{\\frac{\\Pi}{4}}\\left(2\\sec^{2}x+x^{3}+2\\right).dx$","type":"$","id":"4-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1","font":{"color":"#000000","family":"Arial","size":10},"ts":1602147947631,"cs":"DaKCho9sCzDKvMKfKIHSXA==","size":{"width":197,"height":24}}

Solun:- Let f(x) = 2sec2x + x3 + 2

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-1-0-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\left(2\\sec^{2}x+x^{3}+2\\right).dx}\t\n\\end{align*}","type":"align*","ts":1602147986508,"cs":"t2lSWpD7+ZTGH10s1zVp2Q==","size":{"width":206,"height":40}}

We know that:-

{"type":"$","code":"$\\int_{}^{}\\sec^{2}x.dx=\\tan x+C$","id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1602148072744,"cs":"d+C5CHmEFCcXeINpj2DZ9g==","size":{"width":164,"height":18}}

{"id":"3-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","ts":1602148135796,"cs":"/439Pd3xDfUh9usqrvGtbQ==","size":{"width":136,"height":21}}

{"type":"align*","code":"\\begin{align*}\n{I}&={\\left[2\\tan x+\\frac{x^{4}}{4}+2x\\right]_{0}^{\\frac{\\Pi}{4}}}\\\\\n{I}&={\\left[\\left(2\\tan\\frac{\\Pi}{4}+\\frac{\\left(\\frac{\\Pi}{4}\\right)^{4}}{4}+\\frac{2\\Pi}{4}\\right)-\\left(2\\tan0+0+0\\right)\\right]}\\\\\n{I}&={2+\\frac{\\Pi^{4}}{1024}+\\frac{\\Pi}{2}}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1-3-1-0","ts":1602148527749,"cs":"JfieFGmfSsrkRvEfc1TQkw==","size":{"width":366,"height":140}}

{"code":"$18.\\,\\int_{0}^{\\Pi}\\left(\\sin^{2}\\frac{x}{2}-\\cos^{2}\\frac{x}{2}\\right).dx$","id":"4-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-0","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1602149777030,"cs":"0wHK6T01hXDf2FhkbP0jlA==","size":{"width":188,"height":21}}

Solun:- Let f(x) = {"font":{"color":"#000000","size":10,"family":"Arial"},"code":"$\\sin^{2}\\frac{x}{2}-\\cos^{2}\\frac{x}{2}$","type":"$","id":"13-0","ts":1602149820137,"cs":"V9sQR4TymLED/oQjUnbpMw==","size":{"width":104,"height":18}}

Integrate f(x):-

{"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{I}&={\\int_{0}^{\\Pi}\\left(\\sin^{2}\\frac{x}{2}-\\cos^{2}\\frac{x}{2}\\right).dx}\\\\\n{I}&={-\\int_{0}^{\\Pi}\\left(\\cos^{2}\\frac{x}{2}-\\sin^{2}\\frac{x}{2}\\right).dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-1-0-1-1-0-0","ts":1602149880698,"cs":"SNu96LvmYfcKbyIxJymnbQ==","size":{"width":222,"height":82}}

We know that:-

cos 2x = cos2x - sin2x

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-1-0-1-1-1-0","type":"align*","code":"\\begin{align*}\n{I}&={-\\int_{0}^{\\Pi}\\cos x.dx}\t\n\\end{align*}","ts":1602150004015,"cs":"5T8L8Pt1TY74WVkRkVIUeQ==","size":{"width":128,"height":38}}

We know that:-

{"type":"$","code":"$\\int_{}^{}\\cos x.dx=\\sin x+C$","font":{"size":10,"color":"#000000","family":"Arial"},"id":"14-0","ts":1602150063481,"cs":"nHQ/ffN6EOyGF2SzKf/qow==","size":{"width":154,"height":17}}

{"code":"\\begin{align*}\n{I}&={-\\left[\\sin x\\right]_{0}^{\\Pi}}\\\\\n{I}&={-\\left[\\left(\\sin\\Pi\\right)-\\left(\\sin0\\right)\\right]}\\\\\n{I}&={0}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1-3-1-1-0","type":"align*","ts":1602150204240,"cs":"8TdU8PfyyB74jukRGR/biA==","size":{"width":157,"height":60}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"4-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-0","code":"$19.\\,\\int_{0}^{2}\\frac{6x+3}{x^{2}+4}.dx$","type":"$","ts":1602150638051,"cs":"rVPXCUYE2SFP5BRrPl2brg==","size":{"width":102,"height":22}}

Solun:- Let f(x) = {"code":"$\\frac{6x+3}{x^{2}+4}$","type":"$","id":"13-1-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1602150716805,"cs":"pFy0jJV/W7YjMX6JPj++8Q==","size":{"width":32,"height":20}}

Integrate f(x):-

{"code":"\\begin{align*}\n{I}&={\\int_{0}^{2}\\frac{6x+3}{x^{2}+4}.dx}\\\\\n{I}&={\\int_{0}^{2}\\frac{6x}{x^{2}+4}.dx+\\int_{0}^{2}\\frac{3}{x^{2}+4}.dx}\\\\\n{I}&={3\\int_{0}^{2}\\frac{2x}{x^{2}+4}.dx+3\\int_{0}^{2}\\frac{1}{x^{2}+\\left(2\\right)^{2}}.dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-1-0-1-1-0-1-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1602151149811,"cs":"6VRE/RYCYJrudVy8bZvyqA==","size":{"width":286,"height":130}}

We know that:-

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"15-0-0","type":"$","code":"$\\int_{}^{}\\frac{f^{\\prime }\\left(x\\right)}{f\\left(x\\right)}dx=\\log_{}\\left|f\\left(x\\right)\\right|+C$","ts":1602151337311,"cs":"yWxgLHgJHslfOYWDavaGZA==","size":{"width":177,"height":24}}

{"type":"$","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}\\frac{1}{x^{2}+a^{2}}dx=\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+C$","id":"15-1-0","ts":1602151395867,"cs":"nBD6nFO3tuPhSVYK5fLBHA==","size":{"width":201,"height":20}}

{"code":"\\begin{align*}\n{I}&={3\\left[\\log_{}\\left|x^{2}+4\\right|+\\frac{1}{2}\\tan^{-1}\\left(\\frac{x}{2}\\right)\\right]_{0}^{2}}\\\\\n{I}&={3\\left[\\left(\\log_{}\\left|8\\right|+\\frac{1}{2}\\tan^{-1}\\left(1\\right)\\right)-\\left(\\log_{}\\left|4\\right|+\\frac{1}{2}\\tan^{-1}\\left(\\frac{0}{2}\\right)\\right)\\right]}\\\\\n{I}&={3\\left[\\log_{}\\left|8\\right|+\\frac{1}{2}.\\frac{\\Pi}{4}-\\log_{}\\left|4\\right|\\right]}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1-3-1-1-1-0-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","ts":1602152088691,"cs":"3iu4CONBCAmvGg55A9eVGw==","size":{"width":408,"height":126}}

We know that:-

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

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1-3-1-1-1-1","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={3\\left[\\log_{}\\left|\\frac{8}{4}\\right|+\\frac{\\Pi}{8}\\right]}\\\\\n{I}&={3\\log_{}\\left|2\\right|+\\frac{3\\Pi}{8}}\t\n\\end{align*}","type":"align*","ts":1602152218004,"cs":"ih/Thca7f23nj+q9YQ0G+Q==","size":{"width":140,"height":74}}

{"code":"$20.\\,\\int_{0}^{1}\\left(x.e^{x}+\\sin\\frac{\\Pi x}{4}\\right).dx$","id":"4-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","ts":1602152464891,"cs":"k0A3OOt4HF7ArSI6Qj6OTw==","size":{"width":174,"height":21}}

Solun:- Let f(x) = {"id":"13-1-1-0","type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\left(x.e^{x}+\\sin\\frac{\\Pi x}{4}\\right)$","ts":1602152505310,"cs":"zPAgcri4g2dAeXVQIvtZZw==","size":{"width":108,"height":18}}

Integrate f(x):-

{"font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-1-0-1-1-0-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{0}^{1}\\left(x.e^{x}+\\sin\\frac{\\Pi x}{4}\\right).dx}\t\n\\end{align*}","type":"align*","ts":1602152552623,"cs":"Vt9rGekyWxVE22E/bWCbCw==","size":{"width":198,"height":40}}

We know that:-

{"font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}\\left(u.v\\right).dx=u\\int_{}^{}v.dx-\\int_{}^{}\\left[\\diff{u}{x}\\int_{}^{}v.dx\\right].dx+C$","id":"15-0-1-0","type":"$","ts":1602152690602,"cs":"0q0Cpo2s7vV9esp3rEh6cw==","size":{"width":318,"height":20}}

{"font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\int_{}^{}\\sin x.dx=-\\cos x+C$","id":"15-1-1-0","type":"$","ts":1602152715737,"cs":"vmsldaYQiV2LArQR0I40hQ==","size":{"width":169,"height":17}}

{"code":"\\begin{align*}\n{I}&={\\left[x\\int_{}^{}e^{x}.dx-\\int_{}^{}\\left[\\diff{\\left(x\\right)}{x}\\int_{}^{}e^{x}.dx\\right].dx\\right]-\\left[\\frac{\\cos\\frac{\\Pi x}{4}}{\\frac{\\Pi }{4}}\\right]_{0}^{1}}\\\\\n{I}&={\\left[x.e^{x}-\\int_{}^{}e^{x}.dx\\right]-\\frac{4}{\\Pi}\\left[\\cos\\frac{\\Pi x}{4}\\right]_{0}^{1}}\\\\\n{I}&={\\left[x.e^{x}-e^{x}\\right]_{0}^{1}-\\frac{4}{\\Pi}\\left[\\cos\\frac{\\Pi x}{4}\\right]_{0}^{1}}\\\\\n{I}&={\\left[\\left(e-e\\right)-\\left(0-e^{0}\\right)\\right]-\\frac{4}{\\Pi}\\left[\\left(\\cos\\frac{\\Pi}{4}\\right)-\\left(\\cos0\\right)\\right]}\\\\\n{I}&={1-\\frac{4}{\\Pi}\\left(\\frac{1}{{\\sqrt[]{2}}}-1\\right)}\\\\\n{I}&={1-\\frac{2{\\sqrt[]{2}}}{\\Pi}+\\frac{4}{\\Pi}}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1-3-1-1-1-0-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1602153306555,"cs":"+EhGW+7NVMq7vY24il88QA==","size":{"width":388,"height":272}}

Choose the correct answer in Exercises 21 and 22.

{"code":"$21.\\,\\int_{1}^{{\\sqrt[]{3}}}\\frac{dx}{1+x^{2}}$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","id":"4-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-0","ts":1602153530142,"cs":"gSAVYpN7vmmOUd358ToLMQ==","size":{"width":86,"height":25}}

Solun:- Let f(x) = {"font":{"family":"Arial","color":"#000000","size":10},"type":"$","id":"13-1-1-1-0","code":"$\\frac{1}{1+x^{2}}$","ts":1602153566531,"cs":"ax6q8CVJ2Jjb2s15bA1yyg==","size":{"width":32,"height":20}}

Integrate f(x):-

{"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{1}^{{\\sqrt[]{3}}}\\frac{dx}{1+x^{2}}}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-1-0-1-1-0-1-1-1-0","ts":1602153609365,"cs":"apmcFkQGuADnI3dqLgzx8g==","size":{"width":112,"height":41}}

We know that:-

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","id":"15-1-1-1-0","code":"$\\int_{}^{}\\frac{1}{1+x^{2}}.dx=\\tan^{-1}x+C$","ts":1602153673681,"cs":"aVrjAXozWTdDs2t7EuQRBg==","size":{"width":172,"height":20}}

{"font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1-3-1-1-1-0-1-1-0","code":"\\begin{align*}\n{I}&={\\left[\\tan^{-1}x\\right]_{1}^{{\\sqrt[]{3}}}}\\\\\n{I}&={\\left[\\left(\\tan^{-1}\\left({\\sqrt[]{3}}\\right)\\right)-\\left(\\tan^{-1}\\left(1\\right)\\right)\\right]}\\\\\n{I}&={\\frac{\\Pi}{3}-\\frac{\\Pi}{4}}\\\\\n{I}&={\\frac{\\Pi}{12}}\t\n\\end{align*}","type":"align*","ts":1602153941506,"cs":"5pOi/k5yX+fLyayZ5sLs2Q==","size":{"width":238,"height":133}}

The correct answer is D.

{"font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","id":"4-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1","code":"$22.\\,\\int_{0}^{\\frac{2}{3}}\\frac{dx}{4+9x^{2}}$","ts":1602154178692,"cs":"oxHwx+jWa1dRoiP4Ji+Wig==","size":{"width":85,"height":25}}

Solun:- Let f(x) = {"id":"13-1-1-1-1","type":"$","font":{"color":"#000000","family":"Arial","size":10},"code":"$\\frac{1}{4+9x^{2}}$","ts":1602154205066,"cs":"7exiYvPX3plSiJvag9aZSQ==","size":{"width":37,"height":20}}

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-0-1-1-1-0-1-1-0-1-1-1-1","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{2}{3}}\\frac{dx}{4+9x^{2}}}\\\\\n{I}&={\\frac{1}{9}\\int_{0}^{\\frac{2}{3}}\\frac{dx}{\\frac{4}{9}+x^{2}}}\\\\\n{I}&={\\frac{1}{9}\\int_{0}^{\\frac{2}{3}}\\frac{dx}{\\left(\\frac{2}{3}\\right)^{2}+x^{2}}}\t\n\\end{align*}","ts":1602154356885,"cs":"1lsoBFBmA3alRAZR1xCeIQ==","size":{"width":149,"height":145}}

We know that:-

{"id":"15-1-1-1-1","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$\\int_{}^{}\\frac{1}{a^{2}+x^{2}}.dx=\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+C$","type":"$","ts":1602154470192,"cs":"jTLsbZrJo0SIHHKbCqtTUw==","size":{"width":208,"height":20}}

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1-3-1-1-1-0-1-1-1","type":"align*","code":"\\begin{align*}\n{I}&={\\frac{1}{9}\\left[\\frac{1}{\\frac{2}{3}}\\tan^{-1}\\left(\\frac{x}{\\frac{2}{3}}\\right)\\right]_{0}^{\\frac{2}{3}}}\\\\\n{I}&={\\frac{1}{9}\\left[\\frac{3}{2}\\tan^{-1}\\left(\\frac{3x}{2}\\right)\\right]_{0}^{\\frac{2}{3}}}\\\\\n{I}&={\\frac{3}{18}\\left[\\tan^{-1}\\left(\\frac{3x}{2}\\right)\\right]_{0}^{\\frac{2}{3}}}\\\\\n{I}&={\\frac{1}{6}\\left[\\left(\\tan^{-1}\\left(1\\right)\\right)-\\left(\\tan^{-1}\\left(0\\right)\\right)\\right]}\\\\\n{I}&={\\frac{1}{6}.\\frac{\\Pi}{4}}\\\\\n{I}&={\\frac{\\Pi}{24}}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1602154878077,"cs":"8lXk9E+IT7ARCjKSAxZkyw==","size":{"width":226,"height":262}}

The correct answer is C.


Download PDF of Exercise 7.9

See also:-

Notes of Integrals

Exercise 7.1

Exercise 7.2

Exercise 7.3

Exercise 7.4

Exercise 7.5

Exercise 7.6

Exercise 7.7


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