$\smallint \,\,{\tan ^{ - 1}}\,\,x\,\,dx = ...\,\, + c$

$\frac{1}{{1\,\, + \,\,{x^2}}}$

$x\,\,{\tan ^{ - 1}}\,\,x\,\, + \,\,\frac{1}{2}\,\,.\,\,\frac{{{{\tan }^{ - 1}}\,\,x}}{{1 + {x^2}}}$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1

\int \,\,{\tan ^{ - 1}}\,\,x\,\,dx = ...\,\, + c

View answer and explanation

Correct option: D

x\,\,{\tan ^{ - 1}}\,\,x\,\, - \,\,\frac{1}{2}\,\,\log \,\,\left| {1\,\, + \,\,{x^2}} \right|

Option D (

x\,\tan ^{ - 1}\,x\, - \,\frac{1}{2}\,\log \,\left| {1\, + \,{x^2}} \right|

) is correct because it is the result of integrating

\tan^{-1}x

using integration by parts.