Integrals — Maths STD 12 Science — Question

$= \int {\left( {{{\tan }^{ - 1}}x} \right).1} dx$

$ = {\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} $

$= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left| {\left( {1 + {x^2}} \right)} \right| + c$

$\left[ {\because \int {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx = \log \left| {f\left( x \right)} \right|} } \right]$

$= x{\tan ^{ - 1}}x - \frac{1}{2}\log \left( {1 + {x^2}} \right) + c$