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Exercise 7.6

Integrate the functions in Exercises 1 to 22.

1. x.sinx

Solun:- Let f(x) = x.sinx

Integrate f(x):-

{"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}x.\\sin x.dx$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-0-0","ts":1601626122804,"cs":"aNS8IIPEhV4QI0oiDsYxUQ==","size":{"width":216,"height":22}}

We know that:-

{"type":"$","id":"3-0-0-0-0-0-0-0-0","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"Q1AhhbomJnXf/JNAhWw2Vg==","size":{"width":300,"height":20}}

According to ILATE:-

Let u = x and v = sin x

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\int_{}^{}\\sin x.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}\\sin x.dx\\right).\\diff{x}{x}\\right].dx}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0","ts":1601626558839,"cs":"CkireReP8ZITiT6o2hgMrg==","size":{"width":385,"height":37}}

We know that:-

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"1-0-0","code":"$\\int_{}^{}\\sin x.dx=-\\cos x+c$","type":"$","ts":1601626665566,"cs":"xHMBwZweEDJ5tRcYK3tuEA==","size":{"width":164,"height":17}}

{"type":"align*","id":"3-0-0-0-0-0-0-1-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-x.\\cos x-\\int_{}^{}\\left[\\left(-\\cos x\\right).1\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={-x.\\cos x+\\int_{}^{}\\cos x.dx+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601627146696,"cs":"Fa1ydSCs7d6bqn2mOZcOpg==","size":{"width":322,"height":76}}

We know that:- 

{"type":"$","code":"$\\int_{}^{}\\cos x.dx=\\sin x+c$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"1-1","ts":1601627050332,"cs":"aix5rVEclEUmqyxqd+D3Jw==","size":{"width":149,"height":17}}

{"id":"3-0-0-0-0-0-0-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-x.\\cos x+\\sin x+C}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"ts":1601627161966,"cs":"0rlo8yB7I7XLXq5PMmF6Fg==","size":{"width":230,"height":36}}

2. x.sin3x

Solun:- Let f(x) = x.sin3x

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-1-0","type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}x.\\sin 3x.dx$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601802971472,"cs":"iqz5ilcMLlfMQp+x9oX92w==","size":{"width":176,"height":17}}

We know that:-

{"type":"$","id":"3-0-0-0-0-0-0-0-1-0","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"KYnnpSA9ykI0rVfP/sU/ZA==","size":{"width":300,"height":20}}

According to ILATE:-

Let u = x and v = sin 3x 

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\int_{}^{}\\sin 3x.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}\\sin 3x.dx\\right).\\diff{x}{x}\\right].dx}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-0","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"ts":1601627313372,"cs":"26BeOj1FsYn+sNN7J0e+YQ==","size":{"width":401,"height":37}}

We know that:- 

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"1-0-1-0","code":"$\\int_{}^{}\\sin x.dx=-\\cos x+c$","type":"$","ts":1601626665566,"cs":"RRrPrbKfTk5W9ygvnwsfVw==","size":{"width":164,"height":17}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-x.\\frac{\\cos 3x}{3}-\\int_{}^{}\\left[\\frac{\\left(-\\cos 3x\\right)}{3}.1\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\frac{1}{3}x.\\cos 3x+\\frac{1}{3}\\int_{}^{}\\cos 3x.dx+C}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"3-0-0-0-0-0-0-1-1-0-1-0-0","ts":1601627539829,"cs":"xTmU0jlUganvLoUrdG8lxg==","size":{"width":360,"height":78}}

We know that:- 

{"type":"$","code":"$\\int_{}^{}\\cos x.dx=\\sin x+c$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"1-1","ts":1601627050332,"cs":"aix5rVEclEUmqyxqd+D3Jw==","size":{"width":149,"height":17}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\frac{1}{3}x.\\cos 3x+\\frac{1}{3}.\\frac{\\sin3x}{3}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\frac{x}{3}\\cos 3x+\\frac{1}{9}\\sin3x+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-1-0-1-1","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601803423217,"cs":"8ahIl4sTNF8xzQk5hG7GmA==","size":{"width":290,"height":76}}

3. x2.ex

Solun:- Let f(x) = x2.ex

Integrate f(x):-

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}x^{2}.e^{x}.dx$","id":"2-0-0-0-0-0-0-0-0-0-1-1-0","ts":1601627766296,"cs":"4BzTbmgF20HuzUMi9wuTjg==","size":{"width":157,"height":18}}

We know that:-

{"type":"$","id":"3-0-0-0-0-0-0-0-1-1","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"Iem8ZK+f169qDmFZ041Rcg==","size":{"width":300,"height":20}}

According to ILATE:-

Let u = x2 and v = ex 

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x^{2}.\\int_{}^{}e^{x}.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}e^{x}.dx\\right).\\diff{\\left(x^{2}\\right)}{x}\\right].dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-0","ts":1601627878102,"cs":"YbRtOjPCXYPOMA/Ha6jsMw==","size":{"width":381,"height":46}}

We know that:- 

{"id":"1-0-1-1-0","type":"$","code":"$\\int_{}^{}e^{x}.dx=e^{x}+c$","font":{"color":"#000000","family":"Arial","size":10},"ts":1601627906931,"cs":"+Le/mDx0qsMnYc1ZHIfpTw==","size":{"width":114,"height":17}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-1-0-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x^{2}.e^{x}-\\int_{}^{}\\left[\\left(e^{x}\\right).2x\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x^{2}.e^{x}-2\\int_{}^{}xe^{x}.dx+C}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601628634332,"cs":"AwiSrJrdGxOtfxxDwJC7Ag==","size":{"width":273,"height":76}}

Again let u = x and v = ex

{"id":"3-0-0-0-0-0-0-1-0-1-1-1-1","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x^{2}.e^{x}-2\\left[x\\int_{}^{}e^{x}.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}e^{x}.dx\\right).\\diff{x}{x}\\right].dx\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x^{2}.e^{x}-2\\left[x.e^{x}-\\int_{}^{}\\left(e^{x}\\right).1.dx\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x^{2}.e^{x}-2\\left[x.e^{x}-\\int_{}^{}e^{x}.dx\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x^{2}.e^{x}-2\\left[x.e^{x}-e^{x}\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={e^{x}\\left[x^{2}-2x+2\\right]+C}\t\n\\end{align*}","type":"align*","ts":1601628804816,"cs":"9SW6eWunrdwhxEmV6CVotQ==","size":{"width":460,"height":204}}

4. x.log x

Solun:- Let f(x) = x.log x

Integrate f(x):-

{"type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}x.\\log_{}x.dx$","id":"2-0-0-0-0-0-0-0-0-0-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1601629399077,"cs":"CFj1PMD9tVzS4xCyjuRtVg==","size":{"width":169,"height":17}}

We know that:-

{"type":"$","id":"3-0-0-0-0-0-0-0-1-2","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"VEN0auY+H/CV3W33E9yOWQ==","size":{"width":300,"height":20}}

According to ILATE:-

Let u = log x and v = x 

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}x.\\int_{}^{}x.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}x.dx\\right).\\diff{\\left(\\log_{}x\\right)}{x}\\right].dx}\t\n\\end{align*}","ts":1601629530301,"cs":"enmhKqqGOq6SM36ucF6rjA==","size":{"width":401,"height":37}}

We know that:- 

{"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","type":"$","id":"1-0-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601629598122,"cs":"u8ZeVAbocWquCvYH4VIMQA==","size":{"width":130,"height":21}}

{"type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"id":"3","code":"$\\diff{\\left(\\log_{}x\\right)}{x}=\\frac{1}{x}$","ts":1601629656246,"cs":"4QakrfGV2DrbRO7Lz+Mu3g==","size":{"width":101,"height":28}}

{"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\log_{}x}{2}-\\int_{}^{}\\left[\\left(\\frac{x^{2}}{2}\\right).\\frac{1}{x}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\log_{}x}{2}-\\int_{}^{}\\frac{x}{2}.dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\log_{}x}{2}-\\frac{1}{2}\\int_{}^{}x.dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\log_{}x}{2}-\\frac{1}{2}.\\frac{x^{2}}{2}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\log_{}x}{2}-\\frac{x^{2}}{4}+C}\t\n\\end{align*}","ts":1601631752992,"cs":"hlw2p+CHb6supijSuj/2zA==","size":{"width":328,"height":212}}

5. x.log 2x

Solun:- Let f(x) = x.log 2x

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}x.\\log_{}2x.dx$","type":"$","ts":1601630054262,"cs":"CowEWUh1Z/kmtoe22ArblQ==","size":{"width":177,"height":17}}

We know that:-

{"type":"$","id":"3-0-0-0-0-0-0-0-1-3","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"Aw4GTDWHYHPLbXDYIqB+dA==","size":{"width":300,"height":20}}

According to ILATE:-

Let u = log 2x and v = x

{"font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}2x.\\int_{}^{}x.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}x.dx\\right).\\diff{\\left(\\log_{}2x\\right)}{x}\\right].dx}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-0","ts":1601630140243,"cs":"tjc9v7B7pQoN+l5Z/tdLiA==","size":{"width":417,"height":37}}

We know that:- 

{"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","type":"$","id":"1-0-1-1-1-1","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601629598122,"cs":"T6dde5ApFLLlbw3IZmO5gA==","size":{"width":130,"height":21}}

{"type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"id":"4","code":"$\\diff{\\left(\\log_{}x\\right)}{x}=\\frac{1}{x}$","ts":1601629656246,"cs":"jdFy4UeuyD1epIHkdAVpUQ==","size":{"width":101,"height":28}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-0","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\log_{}2x}{2}-\\int_{}^{}\\left[\\left(\\frac{x^{2}}{2}\\right).\\frac{2}{2x}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\log_{}2x}{2}-\\int_{}^{}\\frac{x}{2}.dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\log_{}2x}{2}-\\frac{1}{2}\\int_{}^{}x.dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\log_{}2x}{2}-\\frac{1}{2}.\\frac{x^{2}}{2}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\log_{}2x}{2}-\\frac{x^{2}}{4}+C}\t\n\\end{align*}","ts":1601631718633,"cs":"DAK7jbqrU+f0nFDsJv7aGw==","size":{"width":342,"height":212}}

6. x2.log x

Solun:- Let f(x) = x2.log x

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-0","code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}x^{2}.\\log_{}x.dx$","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1601631260381,"cs":"t3j0iKMs6lwMfhx5kPenoQ==","size":{"width":176,"height":18}}

We know that:-

{"type":"$","id":"3-0-0-0-0-0-0-0-1-4","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"Z72lxtCHJ073FyVYblEpVQ==","size":{"width":300,"height":20}}

According to ILATE:-

Let u = log x and v = x2

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}x.\\int_{}^{}x^{2}.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}x^{2}.dx\\right).\\diff{\\left(\\log_{}x\\right)}{x}\\right].dx}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"ts":1601631317035,"cs":"5PjgETp3fot8bqb7n/NqZA==","size":{"width":414,"height":37}}

We know that:- 

{"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","type":"$","id":"1-0-1-1-1-2","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601629598122,"cs":"40tVymVgH8O8xbv7+aQQrw==","size":{"width":130,"height":21}}

{"type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"id":"5","code":"$\\diff{\\left(\\log_{}x\\right)}{x}=\\frac{1}{x}$","ts":1601629656246,"cs":"48lGnfu3c7+C2omNgRtzAQ==","size":{"width":101,"height":28}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{3}.\\log_{}x}{3}-\\int_{}^{}\\left[\\left(\\frac{x^{3}}{3}\\right).\\frac{1}{x}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{3}.\\log_{}x}{3}-\\int_{}^{}\\frac{x^{2}}{3}.dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{3}.\\log_{}x}{3}-\\frac{1}{3}\\int_{}^{}x^{2}.dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{3}.\\log_{}x}{3}-\\frac{1}{3}.\\frac{x^{3}}{3}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{3}.\\log_{}x}{3}-\\frac{x^{3}}{9}+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-0","ts":1601631685625,"cs":"15PYJ863JyQItm2SAd2hwA==","size":{"width":328,"height":212}}

7. x.sin-1x

Solun:- Let f(x) = x.sin-1x

Integrate f(x):-

{"type":"$","id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}x.\\sin^{-1}x.dx$","ts":1601631984980,"cs":"KUhvhBax70C0YGQcfzhxxw==","size":{"width":182,"height":18}}

We know that:-

{"type":"$","id":"3-0-0-0-0-0-0-0-1-5","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"Yd2OtWpFdN9Dta0y5d8U0g==","size":{"width":300,"height":20}}

According to ILATE:-

Let u = sin-1x and v = x 

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\sin^{-1}x.\\int_{}^{}x.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}x.dx\\right).\\diff{\\left(\\sin^{-1}x\\right)}{x}\\right].dx}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601632144407,"cs":"+GC2FePBqZsOps8P48YvJQ==","size":{"width":432,"height":46}}

We know that:- 

{"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","type":"$","id":"1-0-1-1-1-3-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601629598122,"cs":"seHysagBGBnChX96/Wq2QQ==","size":{"width":130,"height":21}}

{"font":{"color":"#000000","family":"Arial","size":12},"code":"$\\diff{\\left(\\sin^{-1}x\\right)}{x}=\\frac{1}{{\\sqrt[]{1-x^{2}}}}$","id":"6-0","type":"$","ts":1601632245123,"cs":"Qw5vs0AzaSfLKtLsAaGnHQ==","size":{"width":152,"height":36}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-0-0","font":{"family":"Arial","color":"#000000","size":8},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\sin^{-1}x}{2}-\\int_{}^{}\\left[\\left(\\frac{x^{2}}{2}\\right).\\frac{1}{{\\sqrt[]{1-x^{2}}}}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\sin^{-1}x}{2}+\\frac{1}{2}\\int_{}^{}\\left[\\frac{-x^{2}}{{\\sqrt[]{1-x^{2}}}}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\sin^{-1}x}{2}+\\frac{1}{2}\\int_{}^{}\\left(\\frac{1-1-x^{2}}{{\\sqrt[]{1-x^{2}}}}\\right).dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\sin^{-1}x}{2}+\\frac{1}{2}\\left[\\int_{}^{}\\left(\\frac{1-x^{2}}{{\\sqrt[]{1-x^{2}}}}\\right).dx-\\int_{}^{}\\frac{1}{{\\sqrt[]{1-x^{2}}}}.dx\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\sin^{-1}x}{2}+\\frac{1}{2}\\left[\\int_{}^{}{\\sqrt[]{1-x^{2}}}.dx-\\int_{}^{}\\frac{1}{{\\sqrt[]{1-x^{2}}}}.dx\\right]+C}\t\n\\end{align*}","type":"align*","ts":1601632998420,"cs":"kyWVZFsyeOr/ZhX7eEDWOg==","size":{"width":428,"height":189}}

We know that:- 

{"id":"1-0-1-1-1-4-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","code":"$\\int_{}^{}{\\sqrt[]{a^{2}-x^{2}}}.dx=\\frac{x}{2}{\\sqrt[]{a^{2}-x^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{x}{a}\\right)+c$","ts":1601633245462,"cs":"liC0QCFIACL+J5F2AgjGCA==","size":{"width":326,"height":20}}

{"type":"$","code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{1-x^{2}}}}dx=\\sin^{-1}x+c$","id":"7-0-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601633347927,"cs":"gYsMKX+g8lUWGOY0pSFTRw==","size":{"width":166,"height":22}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-1-0","font":{"size":8,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\sin^{-1}x}{2}+\\frac{1}{2}\\left[\\frac{x}{2}{\\sqrt[]{1-x^{2}}}+\\frac{1}{2}\\sin^{-1}x-\\sin^{-1}x\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\sin^{-1}x}{2}+\\frac{1}{2}\\left[\\frac{x}{2}{\\sqrt[]{1-x^{2}}}-\\frac{1}{2}\\sin^{-1}x\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\sin^{-1}x}{2}+\\frac{1}{4}\\left[x{\\sqrt[]{1-x^{2}}}-\\sin^{-1}x\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\sin^{-1}x}{2}+\\frac{x{\\sqrt[]{1-x^{2}}}}{4}-\\frac{\\sin^{-1}x}{4}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2x^{2}.\\sin^{-1}x-\\sin^{-1}x}{4}+\\frac{x{\\sqrt[]{1-x^{2}}}}{4}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\left(2x^{2}-1\\right)\\sin^{-1}x+\\frac{x{\\sqrt[]{1-x^{2}}}}{4}+C}\t\n\\end{align*}","ts":1601633786984,"cs":"fPdkxWIFtymfTMUghxJ1oA==","size":{"width":400,"height":224}}

8. x.tan-1x

Solun:- Let f(x) = x.tan-1x

Integrate f(x):-

{"type":"$","id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}x.\\tan^{-1}x.dx$","ts":1601633878018,"cs":"nhEmGxRVw01L4qxwubkWZQ==","size":{"width":186,"height":18}}

We know that:-

{"type":"$","id":"3-0-0-0-0-0-0-0-1-6","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"ZHQHNuLD0Y712kny8/YgMQ==","size":{"width":300,"height":20}}

According to ILATE:-

Let u = tan-1x and v = x 

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\tan^{-1}x.\\int_{}^{}x.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}x.dx\\right).\\diff{\\left(\\tan^{-1}x\\right)}{x}\\right].dx}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1601633931464,"cs":"6WUDjPGuMMc5be/Y5gkiZA==","size":{"width":440,"height":46}}

We know that:- 

{"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","type":"$","id":"1-0-1-1-1-3-1","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601629598122,"cs":"kIDPmiwCki5aqUmIzJFEMQ==","size":{"width":130,"height":21}}

{"font":{"family":"Arial","size":12,"color":"#000000"},"type":"$","id":"6-1-0","code":"$\\diff{\\left(\\tan^{-1}x\\right)}{x}=\\frac{1}{1+x^{2}}$","ts":1601633968743,"cs":"OWCtC3afPgqvcY95UxlpEg==","size":{"width":144,"height":32}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\tan^{-1}x}{2}-\\int_{}^{}\\left[\\left(\\frac{x^{2}}{2}\\right).\\frac{1}{1+x^{2}}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\tan^{-1}x}{2}-\\frac{1}{2}\\int_{}^{}\\left[\\frac{x^{2}}{1+x^{2}}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\tan^{-1}x}{2}-\\frac{1}{2}\\int_{}^{}\\left(\\frac{1-1+x^{2}}{1+x^{2}}\\right).dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\tan^{-1}x}{2}-\\frac{1}{2}\\left[\\int_{}^{}\\left(\\frac{1+x^{2}}{1+x^{2}}\\right).dx-\\int_{}^{}\\frac{1}{1+x^{2}}.dx\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\tan^{-1}x}{2}-\\frac{1}{2}\\left[\\int_{}^{}1.dx-\\int_{}^{}\\frac{1}{1+x^{2}}.dx\\right]+C}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":8},"type":"align*","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-0-1-0","ts":1601634186604,"cs":"8ObD5RGQd9Qk6hxgAuf1vA==","size":{"width":408,"height":182}}

We know that:- 

{"code":"$\\int_{}^{}\\frac{1}{1+x^{2}}dx=\\tan^{-1}x+c$","font":{"color":"#000000","size":10,"family":"Arial"},"id":"7-1-0","type":"$","ts":1601635163584,"cs":"lXNbXQcrWBtcio1mbyXsRA==","size":{"width":160,"height":20}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-1-1-0","font":{"size":8,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\tan^{-1}x}{2}-\\frac{1}{2}\\left[x-\\tan^{-1}x\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\tan^{-1}x}{2}-\\frac{x}{2}+\\frac{\\tan^{-1}x}{2}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\tan^{-1}x+\\tan^{-1}x}{2}-\\frac{x}{2}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left(x^{2}+1\\right)\\tan^{-1}x-\\frac{x}{2}+C}\t\n\\end{align*}","type":"align*","ts":1601635874263,"cs":"AfzGnU5xKfCgNON1SHeGPw==","size":{"width":277,"height":140}}

9. x.cos-1x

Solun:- Let f(x) = x.cos-1x

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}x.\\cos^{-1}x.dx$","ts":1601635980795,"cs":"ALOyCtH/4nJzXcQW+xJixg==","size":{"width":184,"height":18}}

We know that:-

{"type":"$","id":"3-0-0-0-0-0-0-0-1-7","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"k8+MwSa/eg4WQvLSJf4vBg==","size":{"width":300,"height":20}}

According to ILATE:-

Let u = cos-1x and v = x 

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\cos^{-1}x.\\int_{}^{}x.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}x.dx\\right).\\diff{\\left(\\cos^{-1}x\\right)}{x}\\right].dx}\t\n\\end{align*}","ts":1601636023779,"cs":"QWYdNTJjcZQLbAxYg/sLOg==","size":{"width":436,"height":46}}

We know that:- 

{"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","type":"$","id":"1-0-1-1-1-3-2","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601629598122,"cs":"0YGyL0I2ns8YkMu39Wqi9A==","size":{"width":130,"height":21}}

{"font":{"color":"#000000","size":12,"family":"Arial"},"code":"$\\diff{\\left(\\cos^{-1}x\\right)}{x}=\\frac{-1}{{\\sqrt[]{1-x^{2}}}}$","id":"6-1-1-0","type":"$","ts":1601636059898,"cs":"pi4PfKbWyYsVscY9NB6TIw==","size":{"width":154,"height":36}}

{"font":{"size":8,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\cos^{-1}x}{2}-\\int_{}^{}\\left[\\left(\\frac{x^{2}}{2}\\right).\\frac{-1}{{\\sqrt[]{1-x^{2}}}}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\cos^{-1}x}{2}-\\frac{1}{2}\\int_{}^{}\\left[\\frac{-x^{2}}{{\\sqrt[]{1-x^{2}}}}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\cos^{-1}x}{2}-\\frac{1}{2}\\int_{}^{}\\left(\\frac{1-1-x^{2}}{{\\sqrt[]{1-x^{2}}}}\\right).dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\cos^{-1}x}{2}-\\frac{1}{2}\\left[\\int_{}^{}\\left(\\frac{1-x^{2}}{{\\sqrt[]{1-x^{2}}}}\\right).dx-\\int_{}^{}\\frac{1}{{\\sqrt[]{1-x^{2}}}}.dx\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\cos^{-1}x}{2}-\\frac{1}{2}\\left[\\int_{}^{}{\\sqrt[]{1-x^{2}}}.dx-\\int_{}^{}\\frac{1}{{\\sqrt[]{1-x^{2}}}}.dx\\right]+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-0-1-1-0","ts":1601636353382,"cs":"qB+qk10n/WZyy4cSeKwfpA==","size":{"width":429,"height":188}}

We know that:- 

{"id":"1-0-1-1-1-4-1","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","code":"$\\int_{}^{}{\\sqrt[]{a^{2}-x^{2}}}.dx=\\frac{x}{2}{\\sqrt[]{a^{2}-x^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{x}{a}\\right)+c$","ts":1601633245462,"cs":"PQrndyW4chYkMYcntF3s/w==","size":{"width":326,"height":20}}

{"type":"$","code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{1-x^{2}}}}dx=\\sin^{-1}x+c$","id":"7-0-1","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601633347927,"cs":"5NEbjLA3jcoavtEJjBgNhA==","size":{"width":166,"height":22}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\cos^{-1}x}{2}-\\frac{1}{2}\\left[\\frac{x}{2}{\\sqrt[]{1-x^{2}}}+\\frac{1}{2}\\sin^{-1}x-\\sin^{-1}x\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\cos^{-1}x}{2}-\\frac{1}{2}\\left[\\frac{x}{2}{\\sqrt[]{1-x^{2}}}-\\frac{1}{2}\\sin^{-1}x\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\cos^{-1}x}{2}-\\frac{1}{2}\\left[\\frac{x}{2}{\\sqrt[]{1-x^{2}}}-\\frac{1}{2}\\sin^{-1}x\\right]+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":8},"type":"align*","ts":1601636657470,"cs":"rFTWxZYjiM0pfqtyoyLaSw==","size":{"width":401,"height":108}}

We know that:-

{"code":"$\\sin^{-1}x=\\frac{\\Pi}{2}-\\cos^{-1}x$","type":"$","id":"8","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601636726317,"cs":"AcLhDJtRoP9fF/V+YPnCyQ==","size":{"width":149,"height":18}}

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\cos^{-1}x}{2}-\\frac{1}{2}\\left[\\frac{x}{2}{\\sqrt[]{1-x^{2}}}-\\frac{1}{2}\\left(\\frac{\\Pi}{2}-\\cos^{-1}x\\right)\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\cos^{-1}x}{2}-\\frac{1}{2}\\left[\\frac{x}{2}{\\sqrt[]{1-x^{2}}}-\\frac{\\Pi}{4}+\\frac{1}{2}\\cos^{-1}x\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\cos^{-1}x}{2}-\\frac{1}{4}\\cos^{-1}x-\\frac{x}{4}{\\sqrt[]{1-x^{2}}}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2x^{2}.\\cos^{-1}x-\\cos^{-1}x}{4}-\\frac{x}{4}{\\sqrt[]{1-x^{2}}}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\left(2x^{2}-1\\right)\\cos^{-1}x-\\frac{x}{4}{\\sqrt[]{1-x^{2}}}+C}\\\\\n\\end{align*}","font":{"color":"#000000","size":8,"family":"Arial"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-1-1-1-1","ts":1601637125574,"cs":"8mMf5XFFcxNL0vH1eTy8VQ==","size":{"width":400,"height":178}}

10. (sin-1x)2

Solun:- Let f(x) = (sin-1x)2

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\left(\\sin^{-1}x\\right)^{2}\\times1.dx$","ts":1601637298335,"cs":"fWOEsF8/9vZc3RTEVc/tag==","size":{"width":212,"height":21}}

We know that:-

{"type":"$","id":"3-0-0-0-0-0-0-0-1-8","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"OEI9nOYL2v2haM2cq1KarA==","size":{"width":300,"height":20}}

According to ILATE:-

Let u = (sin-1x)2 and v = 1

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\left(\\sin^{-1}x\\right)^{2}.\\int_{}^{}1.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}1.dx\\right).\\diff{\\left(\\left(\\sin^{-1}x\\right)^{2}\\right)}{x}\\right].dx}\t\n\\end{align*}","ts":1601637450206,"cs":"Ws6DceJ6hp6NaDog/lJV/g==","size":{"width":473,"height":64}}

We know that:- 

{"id":"1-0-1-1-1-3-3-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$\\int_{}^{}1.dx=x+c$","ts":1601637479465,"cs":"FRkFrbcIz4Mum6HxBhLjNg==","size":{"width":100,"height":17}}

{"font":{"family":"Arial","color":"#000000","size":12},"id":"6-1-1-1-0","type":"$","code":"$\\diff{\\left(\\sin^{-1}x\\right)}{x}=\\frac{1}{{\\sqrt[]{1-x^{2}}}}$","ts":1601637518417,"cs":"6VoUYyLv7sY2WnwtOp2fwg==","size":{"width":152,"height":36}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-0-1-1-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\left(\\sin^{-1}x\\right)^{2}-\\int_{}^{}\\left[\\left(x\\right).2\\sin^{-1}x\\times\\frac{1}{{\\sqrt[]{1-x^{2}}}}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\left(\\sin^{-1}x\\right)^{2}-2\\int_{}^{}\\left[\\frac{x.\\sin^{-1}x}{{\\sqrt[]{1-x^{2}}}}\\right].dx+C}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1601638039936,"cs":"VEjpDMjpu1wXpw0udd7bUQ==","size":{"width":444,"height":84}}

Let sin-1x = t and x = sin t

Differentiate w.r.t. to t:-

{"code":"$\\frac{1}{{\\sqrt[]{1-x^{2}}}}dx=dt$","type":"$","font":{"family":"Arial","color":"#000000","size":10},"id":"9-0","ts":1601638243537,"cs":"K6ArevjTnBeikCrhz5LrtA==","size":{"width":93,"height":22}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-0-1-1-1-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\left(\\sin^{-1}x\\right)^{2}-2\\int_{}^{}t.\\sin t.dt+C}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601638349434,"cs":"dJd2C6HIxTEjSryY4ozR1A==","size":{"width":313,"height":36}}

Again let u = t and v = sin t

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\left(\\sin^{-1}x\\right)^{2}-2\\left[t\\int_{}^{}\\sin t.dt-\\int_{}^{}\\left[\\left(\\int_{}^{}\\sin t.dt\\right).\\diff{t}{t}\\right].dt\\right]+C}\\\\\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-0-1-1-1-1-1-0","ts":1601638648396,"cs":"rGU7VGbpYQxhSNJmbga6KA==","size":{"width":516,"height":37}}

We know that:- 

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"7-0-2-0","type":"$","code":"$\\int_{}^{}\\sin x.dx=-\\cos x+c$","ts":1601638764016,"cs":"UYD24JFFRfT16nOqkVm3Kg==","size":{"width":164,"height":17}}

{"type":"align*","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-0-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":8},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\left(\\sin^{-1}x\\right)^{2}-2\\left[t\\left(-\\cos t\\right)-\\int_{}^{}\\left[\\left(-\\cos t\\right).1\\right].dt\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\left(\\sin^{-1}x\\right)^{2}-2\\left[-t\\cos t+\\int_{}^{}\\cos t.dt\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\left(\\sin^{-1}x\\right)^{2}+2t\\cos t-2\\sin t+C}\t\n\\end{align*}","ts":1601639368067,"cs":"+nv1giwkrqFNnL9U0SkLdw==","size":{"width":380,"height":104}}

Put the value of t:-

sin t = x then cos t = {"code":"${\\sqrt[]{1-x^{2}}}$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","id":"12","ts":1601639655825,"cs":"BzOm8Nk5Eh5Do0Gqt17KfQ==","size":{"width":56,"height":16}}

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\left(\\sin^{-1}x\\right)^{2}+2{\\sqrt[]{1-x^{2}}}\\sin^{-1}x-2x+C}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-0-1-1-1-1-1-1-1","ts":1601805321998,"cs":"W/l0LmPwjmbbqLww1wKfnw==","size":{"width":373,"height":36}}

{"code":"$11.\\,\\frac{x.\\cos^{-1}x}{{\\sqrt[]{1-x^{2}}}}$","id":"10","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1601639546413,"cs":"bLCJWBzXIim0G6BPDve1dQ==","size":{"width":93,"height":32}}

Solun:- Let f(x) = {"font":{"family":"Arial","size":12,"color":"#000000"},"id":"11-0","type":"$","code":"$\\frac{x.\\cos^{-1}x}{{\\sqrt[]{1-x^{2}}}}$","ts":1601639560086,"cs":"Nl9f7zuFezZ1p4RLSXXNpA==","size":{"width":64,"height":32}}

Integrate f(x):-

{"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\frac{x.\\cos^{-1}x}{{\\sqrt[]{1-x^{2}}}}.dx$","id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","ts":1601639580811,"cs":"hOsuVjsONxV/JMPQinHvEA==","size":{"width":170,"height":24}}

Let cos-1x = t and x = cos t

Differentiate w.r.t. to t:-

{"code":"\\begin{align*}\n{\\frac{-1}{{\\sqrt[]{1-x^{2}}}}dx}&={dt}\\\\\n{\\frac{1}{{\\sqrt[]{1-x^{2}}}}dx}&={-dt}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"9-1","ts":1601643238520,"cs":"V3i+LeWQ2TlovOyNYl+uCw==","size":{"width":128,"height":80}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-0-1-1-1-1-0-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}t.\\cos t.\\left(-dt\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\int_{}^{}t.\\cos t.dt+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601643031342,"cs":"Vgk6dz3XX27ln0l18S04Tw==","size":{"width":226,"height":76}}

We know that:-

{"type":"$","id":"3-0-0-0-0-0-0-0-1-9","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"f8Y6UW9T44CvTTXn4RWCSA==","size":{"width":300,"height":20}}

According to ILATE:-

Let u = t and v = cos t

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\left[t.\\int_{}^{}\\cos t.dt-\\int_{}^{}\\left[\\left(\\int_{}^{}\\cos t.dt\\right).\\diff{\\left(t\\right)}{t}\\right].dt\\right]}\t\n\\end{align*}","ts":1601644963330,"cs":"E4paHjNhitq1aGNrqGBUkQ==","size":{"width":408,"height":37}}

We know that:- 

{"type":"$","id":"1-0-1-1-1-3-3-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","ts":1601643425917,"cs":"5uibT1LH+Iu5spGSeblmsQ==","size":{"width":130,"height":21}}

{"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\left[t.\\sin t-\\int_{}^{}\\left[\\left(\\sin t\\right).1\\right].dt\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\left[t.\\sin t-\\int_{}^{}\\sin t.dt\\right]+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-1-0-0","type":"align*","ts":1601645346509,"cs":"5ITAPDLZuzPThgcRcScY4Q==","size":{"width":306,"height":80}}

We know that:- 

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"7-0-2-1","type":"$","code":"$\\int_{}^{}\\sin x.dx=-\\cos x+c$","ts":1601638764016,"cs":"N9kq/TNAZ09bCBdKSn27vg==","size":{"width":164,"height":17}}

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-1-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\left[t.\\sin t+\\cos t\\right]+C}\t\n\\end{align*}","ts":1601645322340,"cs":"etH755wVj4Z5wEW56GUBXA==","size":{"width":229,"height":36}}

Put the value of t:-

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-1-1-1-1-0-1-1-1-1-1-1-3-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\left[{\\sqrt[]{1-x^{2}}}\\cos^{-1}x+x\\right]+C}\t\n\\end{align*}","type":"align*","ts":1601805632626,"cs":"2tzaw+27RWswV+Y2iD28Jw==","size":{"width":284,"height":36}}

12. x.sec2x

Solun:- Let f(x) = x.sec2x

Integrate f(x):-

{"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}x.\\sec^{2}x.dx$","id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1601711803930,"cs":"2rRsSzU9bnzD2wtGkaqtYQ==","size":{"width":174,"height":18}}

According to ILATE:-

Let u = x and v = sec2x

{"type":"$","id":"3-0-0-0-0-0-0-0-1-10","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"ziQox7oFDcdgSmF+Ujwyzw==","size":{"width":300,"height":20}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\int_{}^{}\\sec^{2}x.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}\\sec^{2}x.dx\\right).\\diff{\\left(x\\right)}{x}\\right].dx}\t\n\\end{align*}","ts":1601712384601,"cs":"CE92JxjjPe9zvuTPlvoPSg==","size":{"width":412,"height":37}}

We know that:- 

{"code":"$\\int_{}^{}\\sec^{2}x.dx=\\tan x+c$","type":"$","id":"14","font":{"family":"Arial","color":"#000000","size":10},"ts":1601712364040,"cs":"Vt18b14NvhbnjS5sqKPDjw==","size":{"width":158,"height":18}}

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\tan x-\\int_{}^{}\\left[\\left(\\tan x\\right).1\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\tan x-\\int_{}^{}\\tan x.dx+C}\t\n\\end{align*}","id":"16","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601712598077,"cs":"shx4EtJfYJh7W0ONhz9NQg==","size":{"width":298,"height":76}}

We know that:- 

{"type":"$","code":"$\\int_{}^{}\\tan x.dx=-\\log_{}\\left|\\cos x\\right|+c$","id":"7-0-2-2","font":{"family":"Arial","color":"#000000","size":10},"ts":1601712642102,"cs":"+kDVz4NprDTU07L5UIcQIw==","size":{"width":200,"height":17}}

{"font":{"size":10,"family":"Arial","color":"#000000"},"id":"17","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\tan x+\\log_{}\\left|\\cos x\\right|+C}\t\n\\end{align*}","ts":1601712680325,"cs":"pTq5lkC8We/aDDqE5xwAew==","size":{"width":254,"height":36}}

13. tan-1x

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

Integrate f(x):-

{"type":"$","id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\tan^{-1}x\\times1.dx$","ts":1601712988282,"cs":"7Ni40FZwyMpz2qyftuQKmA==","size":{"width":194,"height":18}}

According to ILATE:-

Let u = tan-1x and v = 1

{"type":"$","id":"3-0-0-0-0-0-0-0-1-11","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"cR8FTjGWfKAJ8J5OPCzzHA==","size":{"width":300,"height":20}}

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\tan ^{-1}x.\\int_{}^{}1.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}1.dx\\right).\\diff{\\left(\\tan ^{-1}x\\right)}{x}\\right].dx}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-0","ts":1601713188926,"cs":"kttC06PsF4Xd80d2h912MQ==","size":{"width":432,"height":46}}

We know that:- 

{"id":"1-0-1-1-1-3-3-1-1-0-0","type":"$","code":"$\\int_{}^{}1.dx=x+c$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601713262447,"cs":"AgNH7kttH1CgNfIYUAwYtA==","size":{"width":100,"height":17}}

{"id":"13-0","type":"$","font":{"family":"Arial","color":"#000000","size":12},"code":"$\\diff{\\left(\\tan^{-1}x\\right)}{x}=\\frac{1}{1+x^{2}}$","ts":1601713301881,"cs":"YnvQGCtqpXRCLbug2DLUZQ==","size":{"width":144,"height":32}}

{"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\tan ^{-1}x-\\int_{}^{}\\left[\\left(x\\right).\\frac{1}{1+x^{2}}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\tan ^{-1}x-\\int_{}^{}\\left(\\frac{x}{1+x^{2}}\\right).dx+C}\t\n\\end{align*}","type":"align*","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-1-0-1-0-0","ts":1601713474800,"cs":"TrTQSvJpcyG+3rwuJo2GwA==","size":{"width":340,"height":80}}

Let 1+x2 = t

Differentiate w.r.t. to t:-

⇒ 2x.dx = dt

⇒ x.dx = (1/2).dt

{"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\tan ^{-1}x-\\int_{}^{}\\left(\\frac{1}{t}\\right).\\frac{dt}{2}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\tan ^{-1}x-\\frac{1}{2}\\int_{}^{}\\frac{1}{t}.dt+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-1-0-1-0-1","ts":1601713768164,"cs":"tnF0yTA6spo9iyvtuWMlcg==","size":{"width":288,"height":77}}

We know that:- 

{"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}\\left|x\\right|+c$","id":"15","font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","ts":1601713858043,"cs":"mTnRqd9iCsvevgeR4Wu0IA==","size":{"width":140,"height":18}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\tan ^{-1}x-\\frac{1}{2}\\log_{}\\left|t\\right|+C}\t\n\\end{align*}","id":"18-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601713929336,"cs":"mwwirgmldmdQdWGGNFAe6Q==","size":{"width":257,"height":36}}

Put the value of t:-

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x.\\tan ^{-1}x-\\frac{1}{2}\\log_{}\\left|1+x^{2}\\right|+C}\t\n\\end{align*}","type":"align*","id":"18-1","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1601713988013,"cs":"sglk6k0G6ZBUMR22SaeI9Q==","size":{"width":296,"height":36}}

14. x.(log x)2

Solun:- Let f(x) = x.(log x)2

Integrate f(x):-

{"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}x.\\left(\\log_{}x\\right)^{2}.dx$","id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-0","type":"$","font":{"family":"Arial","color":"#000000","size":10},"ts":1601714152648,"cs":"pPTnMkP35CpcNOQIYTJT4Q==","size":{"width":188,"height":20}}

According to ILATE:-

Let u = (log x)2 and v = x

{"type":"$","id":"3-0-0-0-0-0-0-0-1-12","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"fLxn9uVkgXCB0uHOQHKraA==","size":{"width":300,"height":20}}

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\left(\\log_{}x\\right)^{2}.\\int_{}^{}x.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}x.dx\\right).\\diff{\\left(\\log_{}x\\right)^{2}}{x}\\right].dx}\t\n\\end{align*}","ts":1601714297112,"cs":"JRekrOFzbelmPmiohlSSgQ==","size":{"width":428,"height":46}}

We know that:- 

{"id":"1-0-1-1-1-3-3-1-1-0-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","ts":1601714329238,"cs":"1P+oSdIRERKpLAiBG+DRww==","size":{"width":130,"height":21}}

{"id":"13-1","code":"$\\diff{\\left(\\log_{}x\\right)}{x}=\\frac{1}{x}$","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1601714374306,"cs":"t3yBOltFNus9bk3TLbRO/w==","size":{"width":101,"height":28}}

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\left(\\log_{}x\\right)^{2}}{2}-\\int_{}^{}\\left[\\left(\\frac{x^{2}}{2}\\right).2\\log_{}x.\\frac{1}{x}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\left(\\log_{}x\\right)^{2}}{2}-\\int_{}^{}\\left(x.\\log_{}x\\right).dx+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-1-0-1-0-2-0-0","ts":1601806008198,"cs":"kTMloQ9RgVxT49JP/sV40A==","size":{"width":393,"height":85}}

Again let u = log x and v = x

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\left(\\log_{}x\\right)^{2}}{2}-\\left[\\left(\\log_{}x\\right)\\int_{}^{}x.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}x.dx\\right).\\diff{\\left(\\log_{}x\\right)}{x}\\right].dx\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\left(\\log_{}x\\right)^{2}}{2}-\\left[\\frac{x^{2}.\\log_{}x}{2}-\\int_{}^{}\\left[\\left(\\frac{x^{2}}{2}\\right).\\frac{1}{x}\\right].dx\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\left(\\log_{}x\\right)^{2}}{2}-\\left[\\frac{x^{2}.\\log_{}x}{2}-\\frac{1}{2}\\int_{}^{}x.dx\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\left(\\log_{}x\\right)^{2}}{2}-\\left[\\frac{x^{2}.\\log_{}x}{2}-\\frac{1}{2}.\\frac{x^{2}}{2}\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\left(\\log_{}x\\right)^{2}}{2}-\\left[\\frac{x^{2}.\\log_{}x}{2}-\\frac{x^{2}}{4}\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}.\\left(\\log_{}x\\right)^{2}}{2}-\\frac{x^{2}.\\log_{}x}{2}+\\frac{x^{2}}{4}+C}\t\n\\end{align*}","font":{"size":8,"family":"Arial","color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-1-0-1-0-2-1","type":"align*","ts":1601715654578,"cs":"9K5n3O4nHiufaxyjR1KWPQ==","size":{"width":478,"height":232}}

15. (x2+1).log x

Solun:- Let f(x) = (x2+1).log x

Integrate f(x):-

{"type":"$","id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-0","code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\left(x^{2}+1\\right).\\log_{}x.dx$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601715981342,"cs":"zy6zDu8JUmBXeQr6zCpbvA==","size":{"width":214,"height":18}}

According to ILATE:-

Let u = log x and v = x2 + 1

{"type":"$","id":"3-0-0-0-0-0-0-0-1-13","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"bw/EsAlm5ToTEUr+/ZIw5Q==","size":{"width":300,"height":20}}

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}x.\\int_{}^{}\\left(x^{2}+1\\right).dx-\\int_{}^{}\\left[\\left(\\int_{}^{}\\left(x^{2}+1\\right).dx\\right).\\diff{\\left(\\log_{}x\\right)}{x}\\right].dx}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-1-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1601716120827,"cs":"8LONG+ygyqKmk1dB7QBcDw==","size":{"width":492,"height":37}}

We know that:- 

{"id":"1-0-1-1-1-3-3-1-1-0-1-1","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","ts":1601714329238,"cs":"E5OzR0V18VgrguT/ooT7dw==","size":{"width":130,"height":21}}

{"id":"13-1","code":"$\\diff{\\left(\\log_{}x\\right)}{x}=\\frac{1}{x}$","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1601714374306,"cs":"t3yBOltFNus9bk3TLbRO/w==","size":{"width":101,"height":28}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-1-0-1-0-2-0-1","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\left(\\frac{x^{3}}{3}+x\\right)\\log_{}x-\\int_{}^{}\\left[\\left(\\frac{x^{3}}{3}+x\\right).\\frac{1}{x}\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\left(\\frac{x^{3}}{3}+x\\right)\\log_{}x-\\int_{}^{}\\left[\\left(\\frac{x^{2}}{3}+1\\right)\\right].dx+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\left(\\frac{x^{3}}{3}+x\\right)\\log_{}x-\\left[\\frac{1}{3}\\int_{}^{}x^{2}.dx+\\int_{}^{}1.dx\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\left(\\frac{x^{3}}{3}+x\\right)\\log_{}x-\\left[\\frac{1}{3}.\\frac{x^{3}}{3}+x\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\left(\\frac{x^{3}}{3}+x\\right)\\log_{}x-\\frac{x^{3}}{9}-x+C}\t\n\\end{align*}","type":"align*","ts":1601716515294,"cs":"H5KO3XHzm5kQJTX5ZkONqQ==","size":{"width":409,"height":213}}

16. ex.(sin x + cos x)

Solun:- Let f(x) = ex.(sin x + cos x)

Method 1:-

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-0-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$I=\\int_{}^{}e^{x}.\\left(\\sin x+\\cos x\\right).dx$","type":"$","ts":1601716727159,"cs":"xni1EFP8XNs6T8s/MrlPUA==","size":{"width":184,"height":17}}

We know that:-

{"code":"$\\int_{}^{}e^{x}\\left[f\\left(x\\right)+f^{\\prime }\\left(x\\right)\\right].dx=e^{x}.f\\left(x\\right)+c$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"25","type":"$","ts":1601723144803,"cs":"SF+xX/JNaXddizwq79HJeQ==","size":{"width":244,"height":17}}

Here f(x) = sin x

Then f’(x) = cos x

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{I}&={e^{x}.\\sin x+C}\\\\\n\\end{align*}","id":"28-0","ts":1601728355238,"cs":"z0xNIlaO+LQlP8H76lYizQ==","size":{"width":112,"height":16}}

Method 2:-

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$I=\\int_{}^{}e^{x}.\\left(\\sin x+\\cos x\\right).dx$","type":"$","ts":1601716727159,"cs":"wRRgWUI5PgoNVHHmufp67A==","size":{"width":184,"height":17}}

According to ILATE:-

Let u = sin x + cos x and v = ex

{"type":"$","id":"3-0-0-0-0-0-0-0-1-14","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"Rm3eh/U5s6aIEwAZY1QdsQ==","size":{"width":300,"height":20}}

{"type":"align*","code":"\\begin{align*}\n{I}&={\\left(\\sin x+\\cos x\\right).\\int_{}^{}e^{x}.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}e^{x}.dx\\right).\\diff{\\left(\\sin x+\\cos x\\right)}{x}\\right].dx+C}\\\\\n{I}&={e^{x}.\\left(\\sin x+\\cos x\\right)-\\int_{}^{}e^{x}.\\left(\\cos x-\\sin x\\right).dx+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-1-1-0-0-0","font":{"color":"#000000","family":"Arial","size":8},"ts":1601717662880,"cs":"YJHt+tmsSYcUcsZS+GSokQ==","size":{"width":428,"height":66}}

We know that:- 

{"font":{"color":"#000000","family":"Arial","size":10},"type":"$","id":"1-0-1-1-1-3-3-1-1-0-1-2-0","code":"$\\int_{}^{}e^{x}.dx=e^{x}+c$","ts":1601717087722,"cs":"F50d0Fu5TIt60VN4DovFBQ==","size":{"width":114,"height":17}}

Again let u = cos x - sin x and v = ex

{"code":"\\begin{align*}\n{I}&={e^{x}.\\left(\\sin x+\\cos x\\right)-\\left[\\left(\\cos x-\\sin x\\right)\\int_{}^{}e^{x}.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}e^{x}.dx\\right).\\diff{\\left(\\cos x-\\sin x\\right)}{x}\\right].dx\\right]+C}\\\\\n{I}&={e^{x}.\\left(\\sin x+\\cos x\\right)-\\left[e^{x}.\\left(\\cos x-\\sin x\\right)-\\int_{}^{}\\left[\\left(e^{x}\\right).\\left(-\\sin x-\\cos x\\right)\\right].dx\\right]+C}\\\\\n{I}&={e^{x}.\\left(\\sin x+\\cos x\\right)-\\left[e^{x}.\\left(\\cos x-\\sin x\\right)+\\int_{}^{}\\left[e^{x}.\\left(\\sin x+\\cos x\\right)\\right].dx\\right]+C}\\\\\n{I}&={e^{x}.\\left(\\sin x+\\cos x\\right)-e^{x}.\\left(\\cos x-\\sin x\\right)-I+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-1-1-1-0-0","type":"align*","font":{"color":"#000000","family":"Arial","size":8},"ts":1601717609589,"cs":"4Ab1L5+PiG2xsWYOXoMqtQ==","size":{"width":556,"height":124}}

{"type":"align*","code":"\\begin{align*}\n{2I}&={e^{x}.\\left(\\sin x+\\cos x\\right)-e^{x}.\\left(\\cos x-\\sin x\\right)+C}\\\\\n{2I}&={e^{x}.\\left[\\sin x+\\cos x-\\cos x+\\sin x\\right]+C}\t\n\\end{align*}","font":{"size":11,"color":"#000000","family":"Arial"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-1-1-1-1-0","ts":1601717942239,"cs":"JArne7K6E+qcJpcw0lwAMQ==","size":{"width":368,"height":41}}

{"id":"20-0","code":"\\begin{align*}\n{2I}&={e^{x}.\\left[2\\sin x\\right]+C}\\\\\n{I\\,\\,\\,}&={e^{x}.\\sin x+C}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1601717914491,"cs":"NUzSpMBHVs4f6i7S+NhslQ==","size":{"width":140,"height":36}}

{"font":{"family":"Arial","size":12,"color":"#000000"},"type":"$","id":"21-0","code":"$17.\\,\\frac{x.e^{x}}{\\left(1+x\\right)^{2}}$","ts":1601718113871,"cs":"pPgkjzeGqV6CGcbomkaDIw==","size":{"width":81,"height":32}}

Solun:- Let f(x) = {"code":"$\\frac{x.e^{x}}{\\left(1+x\\right)^{2}}$","type":"$","id":"22-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601718251493,"cs":"a9RbopoyIa9a69djFoSxtQ==","size":{"width":40,"height":24}}

Method 1:-

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0-1","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{x.e^{x}}{\\left(1+x\\right)^{2}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(x+1-1\\right).e^{x}}{\\left(1+x\\right)^{2}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(1+x\\right).e^{x}}{\\left(1+x\\right)^{2}}.dx-\\int_{}^{}\\frac{e^{x}}{\\left(1+x\\right)^{2}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{\\left(1+x\\right)}.e^{x}.dx-\\int_{}^{}\\frac{1}{\\left(1+x\\right)^{2}}.e^{x}.dx}\\\\\n{I}&={\\int_{}^{}e^{x}\\left[\\frac{1}{\\left(1+x\\right)}-\\frac{1}{\\left(1+x\\right)^{2}}\\right].dx}\t\n\\end{align*}","type":"align*","ts":1601807573294,"cs":"OH3UvnqqnsiIBMj7hEEGOA==","size":{"width":296,"height":228}}

We know that:-

{"code":"$\\int_{}^{}e^{x}\\left[f\\left(x\\right)+f^{\\prime }\\left(x\\right)\\right].dx=e^{x}.f\\left(x\\right)+c$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"25","type":"$","ts":1601723144803,"cs":"SF+xX/JNaXddizwq79HJeQ==","size":{"width":244,"height":17}}

Here f(x) = 1/1+x

Then {"id":"26-0","type":"$","code":"$f^{\\prime }\\left(x\\right)=\\frac{-1}{\\left(1+x\\right)^{2}}$","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601723716709,"cs":"LYwpiU712FL8xmIPz1zkKA==","size":{"width":96,"height":24}}

{"type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-1-1-0","code":"\\begin{align*}\n{I}&={\\frac{e^{x}}{1+x}}\t\n\\end{align*}","ts":1601723744306,"cs":"0aKbWc1On7jYDITa++TRAQ==","size":{"width":72,"height":33}}

Method 2:-

Integrate f(x):-

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{x.e^{x}}{\\left(1+x\\right)^{2}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(x+1-1\\right).e^{x}}{\\left(1+x\\right)^{2}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(1+x\\right).e^{x}}{\\left(1+x\\right)^{2}}.dx-\\int_{}^{}\\frac{e^{x}}{\\left(1+x\\right)^{2}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{\\left(1+x\\right)}.e^{x}.dx-\\int_{}^{}\\frac{1}{\\left(1+x\\right)^{2}}.e^{x}.dx}\\\\\n{I}&={I_{1}-\\int_{}^{}\\frac{1}{\\left(1+x\\right)^{2}}.e^{x}.dx....\\left(1\\right)}\t\n\\end{align*}","type":"align*","id":"36","ts":1601807605783,"cs":"wKWMJrLzG2z86tvbYf/Epw==","size":{"width":296,"height":221}}

Calculate I1:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"id":"23-0","type":"$","code":"$I_{1}=\\int_{}^{}\\frac{1}{\\left(1+x\\right)}.e^{x}.dx$","ts":1601718702956,"cs":"95INsT3hGobcBEovuqu+eA==","size":{"width":130,"height":21}}

According to ILATE:-

Let u = 1/(1+x) and v = ex

{"type":"$","id":"3-0-0-0-0-0-0-0-1-15","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"UgRMu5cbmy6lpNnfwnrzNQ==","size":{"width":300,"height":20}}

{"code":"\\begin{align*}\n{I_{1}}&={\\frac{1}{1+x}.\\int_{}^{}e^{x}.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}e^{x}.dx\\right).\\diff{\\left(\\frac{1}{1+x}\\right)}{x}\\right].dx+C}\\\\\n{I_{1}}&={\\frac{e^{x}}{1+x}-\\int_{}^{}e^{x}.\\left(\\frac{-1}{\\left(1+x\\right)^{2}}\\right).dx+C}\\\\\n{I_{1}}&={\\frac{e^{x}}{1+x}+\\int_{}^{}\\frac{1}{\\left(1+x\\right)^{2}}.e^{x}.dx+C}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","size":8,"family":"Arial"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-1-1-0-1-0","ts":1601719094245,"cs":"s/J7vw18uJr4/+JaBEdozQ==","size":{"width":342,"height":124}}

Put the value of I1 in eq. 1:-

{"code":"\\begin{align*}\n{I}&={\\frac{e^{x}}{1+x}+\\int_{}^{}\\frac{1}{\\left(1+x\\right)^{2}}.e^{x}.dx-\\int_{}^{}\\frac{1}{\\left(1+x\\right)^{2}}.e^{x}.dx+C}\\\\\n{I}&={\\frac{e^{x}}{1+x}+C}\t\n\\end{align*}","id":"24-0","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601719360938,"cs":"zf+3EZN2Nwf8jEfAvJQjyQ==","size":{"width":396,"height":77}}

{"type":"$","id":"21-1-0","code":"$18.\\,e^{x}\\left(\\frac{1+\\sin x}{1+\\cos x}\\right)$","font":{"color":"#000000","family":"Arial","size":10},"ts":1601719568672,"cs":"Y4OysSR6qQ1zUpDetDmC/A==","size":{"width":96,"height":20}}

Solun:- Let f(x) = {"code":"$e^{x}\\left(\\frac{1+\\sin x}{1+\\cos x}\\right)$","font":{"color":"#000000","family":"Arial","size":10},"type":"$","id":"22-1-0","ts":1601719593798,"cs":"CNCiJ8Yqo20jVGt0O0jJww==","size":{"width":73,"height":20}}

Integrate f(x):-

{"font":{"size":7.337662337662336,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-1-0-0","type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}e^{x}\\left(\\frac{1+\\sin x}{1+\\cos x}\\right).dx}\\\\\n{I}&={\\int_{}^{}e^{x}\\left(\\frac{\\sin^{2}\\frac{x}{2}+\\cos^{2}\\frac{x}{2}+2\\sin \\frac{x}{2}\\cos\\frac{x}{2}}{2\\cos^{2}\\frac{x}{2}}\\right).dx}\\\\\n{I}&={\\frac{1}{2}\\int_{}^{}e^{x}\\left(\\frac{\\left(\\sin\\frac{x}{2}+\\cos\\frac{x}{2}\\right)^{2}}{\\cos^{2}\\frac{x}{2}}\\right).dx}\\\\\n{I}&={\\frac{1}{2}\\int_{}^{}e^{x}\\left(\\tan\\frac{x}{2}+1\\right)^{2}.dx}\\\\\n{I}&={\\frac{1}{2}\\int_{}^{}e^{x}\\left(1+\\tan^{2}\\frac{x}{2}+2\\tan \\frac{x}{2}\\right).dx}\\\\\n{I}&={\\frac{1}{2}\\int_{}^{}e^{x}\\left(\\sec^{2}\\frac{x}{2}+2\\tan \\frac{x}{2}\\right).dx}\\\\\n{I}&={\\int_{}^{}e^{x}\\left(\\frac{1}{2}\\sec^{2}\\frac{x}{2}+\\tan \\frac{x}{2}\\right).dx}\t\n\\end{align*}","ts":1601724078652,"cs":"wURIbOV+jH022Z2Lxg0NWw==","size":{"width":326,"height":188}}

{"font":{"size":7.337662337662336,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-1-0-1","type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}e^{x}\\left(\\frac{1+\\sin x}{1+\\cos x}\\right).dx}\\\\\n{I}&={\\int_{}^{}e^{x}\\left(\\frac{\\sin^{2}\\frac{x}{2}+\\cos^{2}\\frac{x}{2}+2\\sin \\frac{x}{2}\\cos\\frac{x}{2}}{2\\cos^{2}\\frac{x}{2}}\\right).dx}\\\\\n{I}&={\\frac{1}{2}\\int_{}^{}e^{x}\\left(\\frac{\\left(\\sin\\frac{x}{2}+\\cos\\frac{x}{2}\\right)^{2}}{\\cos^{2}\\frac{x}{2}}\\right).dx}\\\\\n{I}&={\\frac{1}{2}\\int_{}^{}e^{x}\\left(\\tan\\frac{x}{2}+1\\right)^{2}.dx}\\\\\n{I}&={\\frac{1}{2}\\int_{}^{}e^{x}\\left(1+\\tan^{2}\\frac{x}{2}+2\\tan \\frac{x}{2}\\right).dx}\\\\\n{I}&={\\frac{1}{2}\\int_{}^{}e^{x}\\left(\\sec^{2}\\frac{x}{2}+2\\tan \\frac{x}{2}\\right).dx}\\\\\n{I}&={\\int_{}^{}e^{x}\\left(\\frac{1}{2}\\sec^{2}\\frac{x}{2}+\\tan \\frac{x}{2}\\right).dx}\t\n\\end{align*}","ts":1601724078652,"cs":"o7DzUCaSFlxDAVKrYtowKA==","size":{"width":326,"height":226}}

We know that:-

{"code":"$\\int_{}^{}e^{x}\\left[f\\left(x\\right)+f^{\\prime }\\left(x\\right)\\right].dx=e^{x}.f\\left(x\\right)+c$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"25","type":"$","ts":1601723144803,"cs":"SF+xX/JNaXddizwq79HJeQ==","size":{"width":244,"height":17}}

Here f(x) = tan (x/2)

Then {"type":"$","font":{"color":"#000000","family":"Arial","size":10},"code":"$f^{\\prime }\\left(x\\right)=\\frac{1}{2}\\sec^{2}\\left(\\frac{x}{2}\\right)$","id":"26-1-1-0","ts":1601723292894,"cs":"YGbB++bhQI5Ot0shf1mtyQ==","size":{"width":125,"height":18}}

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{I}&={e^{x}\\tan \\frac{x}{2}+C}\t\n\\end{align*}","ts":1601728326281,"cs":"XBz+S2lKqkMquWX8Jzt3kw==","size":{"width":118,"height":28}}

{"font":{"color":"#000000","size":10,"family":"Arial"},"code":"$19.\\,e^{x}\\left(\\frac{1}{x}-\\frac{1}{x^{2}}\\right)$","id":"21-1-1-0","type":"$","ts":1601724466716,"cs":"o7Z8mvmBXr8FJd8haMY5VA==","size":{"width":102,"height":20}}

Solun:- Let f(x) = {"font":{"color":"#000000","family":"Arial","size":10},"id":"22-1-1-0","type":"$","code":"$e^{x}\\left(\\frac{1}{x}-\\frac{1}{x^{2}}\\right)$","ts":1601724527869,"cs":"eAgd8x5yyra78krbLp+l6w==","size":{"width":80,"height":20}}

Integrate f(x):-

{"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}e^{x}\\left(\\frac{1}{x}-\\frac{1}{x^{2}}\\right).dx}\\\\\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-1-0-2-0","font":{"size":7.337662337662336,"family":"Arial","color":"#000000"},"ts":1601724585783,"cs":"comAQ7dlyT9UBSpD3XpquA==","size":{"width":144,"height":32}}

We know that:-

{"code":"$\\int_{}^{}e^{x}\\left[f\\left(x\\right)+f^{\\prime }\\left(x\\right)\\right].dx=e^{x}.f\\left(x\\right)+c$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"25","type":"$","ts":1601723144803,"cs":"SF+xX/JNaXddizwq79HJeQ==","size":{"width":244,"height":17}}

Here f(x) = 1/x

Then {"code":"$f^{\\prime }\\left(x\\right)=\\frac{-1}{x^{2}}$","type":"$","id":"26-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601724665239,"cs":"ZglXDS6VwscVZ7GOGbl0vQ==","size":{"width":76,"height":20}}

{"id":"29-0","code":"$I=\\frac{e^{x}}{x}+C$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601728264829,"cs":"NFRzrTxVqfEmA4kqTndElQ==","size":{"width":77,"height":18}}

{"type":"$","code":"$20.\\,\\,\\frac{\\left(x-3\\right)e^{x}}{\\left(x-1\\right)^{3}}$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"21-1-1-1-0","ts":1601728725898,"cs":"SQI90dZhy1NbwKbkHl3h/A==","size":{"width":72,"height":26}}

Solun:- Let f(x) = {"font":{"size":10,"color":"#000000","family":"Arial"},"id":"22-1-1-1-0","code":"$\\frac{\\left(x-3\\right)e^{x}}{\\left(x-1\\right)^{3}}$","type":"$","ts":1601728712345,"cs":"kk74m8lUJF4BFEqNym27DA==","size":{"width":46,"height":26}}

Integrate f(x):-

{"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{\\left(x-3\\right)e^{x}}{\\left(x-1\\right)^{3}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(x-1-2\\right)e^{x}}{\\left(x-1\\right)^{3}}.dx}\\\\\n{I}&={\\int_{}^{}\\left(\\frac{\\left(x-1\\right)e^{x}}{\\left(x-1\\right)^{3}}-\\frac{2e^{x}}{\\left(x-1\\right)^{3}}\\right).dx}\\\\\n{I}&={\\int_{}^{}e^{x}\\left(\\frac{1}{\\left(x-1\\right)^{2}}-\\frac{2}{\\left(x-1\\right)^{3}}\\right).dx}\\\\\n\\end{align*}","font":{"color":"#000000","size":8.229166666666666,"family":"Arial"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-1-1-0-2-1","ts":1601728677842,"cs":"wByYOjAu/jHR2DXiLtkz8A==","size":{"width":256,"height":158}}

We know that:-

{"code":"$\\int_{}^{}e^{x}\\left[f\\left(x\\right)+f^{\\prime }\\left(x\\right)\\right].dx=e^{x}.f\\left(x\\right)+c$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"25","type":"$","ts":1601723144803,"cs":"SF+xX/JNaXddizwq79HJeQ==","size":{"width":244,"height":17}}

Here f(x) = 1/(x-1)2

Then {"id":"26-1-1-1-1","font":{"color":"#000000","family":"Arial","size":10},"type":"$","code":"$f^{\\prime }\\left(x\\right)=\\frac{-2}{\\left(x-1\\right)^{3}}$","ts":1601728832307,"cs":"ohRLm2dsxYV4MYzREoUfIg==","size":{"width":96,"height":24}}

{"type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"id":"29-1","code":"$I=\\frac{e^{x}}{\\left(x-1\\right)^{2}}+C$","ts":1601728862499,"cs":"TxTgbCpphuJ7nQJ24Yr4/w==","size":{"width":100,"height":24}}

{"id":"21-1-1-1-1-0","code":"$21.\\,\\,e^{2x}\\sin x$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1601812343031,"cs":"gUsbV9MKKfs8ArGpTbuvxg==","size":{"width":101,"height":16}}

Solun:- Let f(x) = {"code":"$e^{2x}\\sin x$","type":"$","id":"22-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":12},"ts":1601812412571,"cs":"SjIxzteup5v3BL8GWenUkg==","size":{"width":68,"height":16}}

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-0-2-0","font":{"color":"#000000","family":"Arial","size":10},"code":"$I=\\int_{}^{}e^{2x}.\\sin x.dx$","type":"$","ts":1601729078580,"cs":"fc6RL4QEj2CqktjpeCFEzg==","size":{"width":125,"height":18}}

According to ILATE:-

Let u = sin x and v = e2x

{"type":"$","id":"3-0-0-0-0-0-0-0-1-16","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"qslkK30p8KCXRIKQeioX1A==","size":{"width":300,"height":20}}

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{I}&={\\sin x.\\int_{}^{}e^{2x}.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}e^{2x}.dx\\right).\\diff{\\left(\\sin x\\right)}{x}\\right].dx+C}\\\\\n{I}&={\\frac{e^{2x}}{2}.\\sin x-\\int_{}^{}\\frac{e^{2x}}{2}.\\cos x.dx+C}\\\\\n{I}&={\\frac{1}{2}e^{2x}.\\sin x-\\frac{1}{2}\\int_{}^{}e^{2x}.\\cos x.dx+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-1-1-0-0-1-0-0","ts":1601729669843,"cs":"edLAbTP4yNfYYHl3HnQVjg==","size":{"width":393,"height":121}}

Again let u = cos x and v = e2x

{"type":"align*","code":"\\begin{align*}\n{I}&={\\frac{1}{2}e^{2x}.\\sin x-\\frac{1}{2}\\int_{}^{}e^{2x}.\\cos x.dx+C}\\\\\n{I}&={\\frac{1}{2}e^{2x}.\\sin x-\\frac{1}{2}\\left[\\cos x\\int_{}^{}e^{2x}.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}e^{2x}.dx\\right)\\diff{\\cos x}{x}\\right].dx\\right]+C}\\\\\n{I}&={\\frac{1}{2}e^{2x}.\\sin x-\\frac{1}{2}\\left[\\frac{e^{2x}}{2}\\cos x-\\int_{}^{}\\left[\\left(\\frac{e^{2x}}{2}\\right)\\left(-\\sin x\\right)\\right].dx\\right]+C}\\\\\n{I}&={\\frac{1}{2}e^{2x}.\\sin x-\\frac{1}{4}\\left[e^{2x}\\cos x+\\int_{}^{}\\left[e^{2x}\\sin x\\right].dx\\right]+C}\\\\\n{I}&={\\frac{1}{2}e^{2x}.\\sin x-\\frac{1}{4}\\left[e^{2x}\\cos x+\\int_{}^{}e^{2x}\\sin x.dx\\right]+C}\\\\\n{I}&={\\frac{1}{2}e^{2x}.\\sin x-\\frac{1}{4}e^{2x}\\cos x-\\frac{I}{4}+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-1-1-0-0-1-1-0","ts":1601729847826,"cs":"i+LvN4nPH2+XbNhNlfK6Nw==","size":{"width":500,"height":244}}

{"type":"align*","id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-1-1-1-1-1","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\frac{5I}{4}}&={\\frac{e^{2x}}{4}.\\left(2\\sin x-\\cos x\\right)+C}\\\\\n{5I}&={e^{2x}.\\left(2\\sin x-\\cos x\\right)+C}\\\\\n\\end{align*}","ts":1601729995655,"cs":"aH63aewBNQkyu6sJSPd6Mw==","size":{"width":216,"height":56}}

{"type":"align*","code":"\\begin{align*}\n{I}&={\\frac{e^{2x}}{5}.\\left(2\\sin x-\\cos x\\right)+C}\\\\\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"30-0","ts":1601729975727,"cs":"S4wHqgpdyM8YfIm+hv4zxg==","size":{"width":200,"height":34}}

{"type":"$","font":{"family":"Arial","color":"#000000","size":10},"code":"$22.\\,\\,\\sin^{-1}\\left(\\frac{2x}{1+x^{2}}\\right)$","id":"21-1-1-1-1-1-0","ts":1601730140742,"cs":"OXDLHVlPv19B+h80O6cNOA==","size":{"width":116,"height":28}}

Solun:- Let f(x) = {"id":"22-1-1-1-1-1-0","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\sin^{-1}\\left(\\frac{2x}{1+x^{2}}\\right)$","ts":1601730161841,"cs":"aQnGYK4FCsl/OTvQMMBBIw==","size":{"width":88,"height":28}}

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-0-2-1-0","type":"$","code":"$I=\\int_{}^{}\\sin^{-1}\\left(\\frac{2x}{1+x^{2}}\\right).dx$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1601730195070,"cs":"AKcQIDTsw+hLll5c8JF1sQ==","size":{"width":156,"height":28}}

Put x = tan y....(1)

Differentiate w.r.t. to y:-

dx = sec2y.dy

{"id":"31-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\sin^{-1}\\left(\\frac{2\\tan y}{1+\\tan^{2}y}\\right).\\sec^{2}y.dy}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1601799333501,"cs":"yHmlKRye4Gku2PZKN5y30w==","size":{"width":249,"height":37}}

We know that:-

{"id":"32","code":"$\\sin2y=\\frac{2\\tan y}{1+\\tan^{2}y}$","font":{"color":"#000000","family":"Arial","size":10},"type":"$","ts":1601799419798,"cs":"hcgfV13F6SfnAZUa6X6MIw==","size":{"width":108,"height":24}}

{"id":"31-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\sin^{-1}\\left(\\sin2y\\right).\\sec^{2}y.dy}\\\\\n{I}&={\\int_{}^{}2y.\\sec^{2}y.dy}\\\\\n{I}&={2\\int_{}^{}y.\\sec^{2}y.dy}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1601799485499,"cs":"J0nPcvwh2snizU8mo1ERlg==","size":{"width":202,"height":116}}

According to ILATE:-

Let u = y and v = sec2y

{"type":"$","id":"3-0-0-0-0-0-0-0-1-16","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\int_{}^{}u.v.dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\left(\\int_{}^{}v.dx\\right).\\diff{u}{x}\\right].dx$","ts":1601626583758,"cs":"qslkK30p8KCXRIKQeioX1A==","size":{"width":300,"height":20}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-1-1-0-0-1-0-1-0","code":"\\begin{align*}\n{I}&={2\\left[y.\\int_{}^{}\\sec^{2}y.dy-\\int_{}^{}\\left[\\left(\\int_{}^{}\\sec^{2}y.dy\\right).\\diff{\\left(y\\right)}{y}\\right].dy\\right]+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1601800024384,"cs":"LMzJaCnzQ6xDrVdVJy/DNg==","size":{"width":400,"height":37}}

We know that:- 

{"code":"$\\int_{}^{}\\sec^{2}x.dx=\\tan x+c$","id":"1-0-1-1-1-3-3-1-1-0-1-2-2","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1601799657335,"cs":"aGTNC6zN2beYRumbTKAhFA==","size":{"width":158,"height":18}}

{"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-1-1-0-0-1-0-1-1-0","type":"align*","code":"\\begin{align*}\n{I}&={2\\left[y.\\tan y-\\int_{}^{}\\left[\\left(\\tan y\\right).1\\right].dy\\right]+C}\\\\\n{I}&={2\\left[y.\\tan y-\\int_{}^{}\\tan y.dy\\right]+C}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1601800072714,"cs":"UDSmCSYvRNjCfVWdpJGvQA==","size":{"width":258,"height":80}}

We know that:-

{"code":"$\\int_{}^{}\\tan x.dx=-\\log_{}\\left|\\cos x\\right|+c$","id":"1-0-1-1-1-3-3-1-1-0-1-2-2","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","ts":1601799824691,"cs":"vEuiBfNgHcB7dgorkI1uww==","size":{"width":200,"height":17}}

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"3-0-0-0-0-0-0-1-0-1-1-0-1-0-1-1-1-1-1-1-1-0-1-0-1-0-1-1-0-0-1-0-1-1-1","code":"\\begin{align*}\n{I}&={2\\left[y.\\tan y-\\left(-\\log_{}\\left|\\cos y\\right|\\right)\\right]+C}\\\\\n{I}&={2\\left[y.\\tan y+\\log_{}\\left|\\cos y\\right|\\right]+C}\t\n\\end{align*}","ts":1601800130942,"cs":"sk0yxBLiRLjjW4Ty+KnCDQ==","size":{"width":236,"height":36}}

From Eq. 1:-

tan y = x and y = tan-1x

{"code":"$\\cos y=\\frac{1}{{\\sqrt[]{1+x^{2}}}}$","type":"$","id":"34","font":{"color":"#000000","family":"Arial","size":10},"ts":1601800499559,"cs":"dExq96wkCoDYfF6IdUuFeA==","size":{"width":93,"height":22}}

{"id":"33","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{I}&={2\\left[x.\\tan^{-1}x+\\log_{}\\left|\\frac{1}{{\\sqrt[]{1+x^{2}}}}\\right|\\right]+C}\\\\\n{I}&={2\\left[x.\\tan^{-1}x+\\log_{}\\left|1\\right|-log\\left|{\\sqrt[]{1+x^{2}}}\\right|\\right]+C}\\\\\n{I}&={2\\left[x.\\tan^{-1}x-\\frac{1}{2}log\\left|1+x^{2}\\right|\\right]+C}\\\\\n{I}&={2x.\\tan^{-1}x-log\\left|1+x^{2}\\right|+C}\t\n\\end{align*}","type":"align*","ts":1601809297640,"cs":"YHFNLJ0KUpnYA6vJ0z/2PQ==","size":{"width":317,"height":140}}

Choose the correct answer in Exercises 23 and 24.

{"code":"$23.\\,\\int_{}^{}x^{2}e^{x^{3}}.dx$","id":"21-1-1-1-1-1-1","font":{"family":"Arial","color":"#000000","size":10},"type":"$","ts":1601798220689,"cs":"7oIlir1I1CDE31GXuQzTdQ==","size":{"width":97,"height":20}} equals

Solun:- Let f(x) = {"type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"id":"22-1-1-1-1-1-1-0","code":"$x^{2}e^{x^{3}}$","ts":1601798249662,"cs":"g201bi4zomaqIYXPErxvpg==","size":{"width":34,"height":14}}

Integrate f(x):-

{"font":{"family":"Arial","color":"#000000","size":10},"type":"$","id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-0-2-1-1-0","code":"$I=\\int_{}^{}x^{2}e^{x^{3}}.dx$","ts":1601798354536,"cs":"3ayZpQlSRvCTUZPouKqD7w==","size":{"width":101,"height":20}}

Put x3 = t

Differentiate w.r.t. to t:-

3x2.dx = dt

x2.dx = (1/3).dt

{"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}e^{t}.\\frac{dt}{3}}\\\\\n{I}&={\\frac{1}{3}\\int_{}^{}e^{t}.dt}\t\n\\end{align*}","id":"31-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1601801821632,"cs":"wOwUs93zBV07+FaNCSJzGQ==","size":{"width":98,"height":76}}

{"type":"align*","code":"\\begin{align*}\n{I}&={\\frac{1}{3}e^{t}+C}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"31-1-1","ts":1601801892672,"cs":"/xXXW1oq1GUmxzuZEJzlFw==","size":{"width":88,"height":32}}

Put the value of t:-

{"code":"\\begin{align*}\n{I}&={\\frac{1}{3}e^{x^{3}}+C}\t\n\\end{align*}","id":"31-1-1","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1601801950569,"cs":"Pjn+lvQCO6GxUNrhtMrB3g==","size":{"width":93,"height":32}}

The correct answer is A.

{"type":"$","font":{"family":"Arial","color":"#000000","size":10},"id":"35","code":"$24.\\,\\int_{}^{}e^{x}\\sec x\\left(1+\\tan x\\right).dx$","ts":1601802045486,"cs":"aXu7NItzFNAzwzBQcGLqow==","size":{"width":186,"height":17}}

Solun:- Let f(x) = {"code":"$e^{x}\\sec x\\left(1+\\tan x\\right)$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","id":"22-1-1-1-1-1-1-1","ts":1601802065562,"cs":"M4OZjYNuxtk9bhLCady3bQ==","size":{"width":122,"height":16}}

Integrate f(x):-

{"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}e^{x}\\sec x\\left(1+\\tan x\\right).dx}\\\\\n{I}&={\\int_{}^{}e^{x}\\left(\\sec x+\\sec x.\\tan x\\right).dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0-1-1-1-0-2-1-1-1","ts":1601802123965,"cs":"eZqrPAgeCu4oQfTLArlPQw==","size":{"width":224,"height":76}}

We know that:-

{"code":"$\\int_{}^{}e^{x}\\left[f\\left(x\\right)+f^{\\prime }\\left(x\\right)\\right].dx=e^{x}.f\\left(x\\right)+c$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"25","type":"$","ts":1601723144803,"cs":"SF+xX/JNaXddizwq79HJeQ==","size":{"width":244,"height":17}}

Here f(x) = sec x

Then f’(x) = sec x.tan x

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"28-1","type":"align*","code":"\\begin{align*}\n{I}&={e^{x}.\\sec x+C}\\\\\n\\end{align*}","ts":1601802213193,"cs":"ecZ/WJ5zdNEBvdk58EJIjw==","size":{"width":113,"height":16}}

The correct answer is B.


See also:-

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