Question
Integrate the function $\tan^{-1}x$
✓
Answer
Let $I = \tan^{-1}xdx$
$= \int {\left( {{{\tan }^{ - 1}}x} \right).1} dx$
$ = {\tan ^{ - 1}}x.x - \int {\frac{1}{{1 + {x^2}}}x.dx} $
$= x{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{2x}}{{1 + {x^2}}}dx} $
$= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left| {\left( {1 + {x^2}} \right)} \right| + c$
$\left[ {\because \int {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx = \log \left| {f\left( x \right)} \right|} } \right]$
$= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left( {1 + {x^2}} \right) + c$
$= \int {\left( {{{\tan }^{ - 1}}x} \right).1} dx$
$ = {\tan ^{ - 1}}x.x - \int {\frac{1}{{1 + {x^2}}}x.dx} $
$= x{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{2x}}{{1 + {x^2}}}dx} $
$= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left| {\left( {1 + {x^2}} \right)} \right| + c$
$\left[ {\because \int {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx = \log \left| {f\left( x \right)} \right|} } \right]$
$= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left( {1 + {x^2}} \right) + c$
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