Integrals — Maths STD 12 Science — Question

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

$= \left( {{{\tan }^{ - 1}}x} \right).\frac{{{x^2}}}{2} - \int {\frac{1}{{1 + {x^2}}}.\frac{{{x^2}}}{2}dx} $

$= \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{{x^2}}}{{1 + {x^2}}}dx} $

$= \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{{x^2} + 1 - 1}}{{{x^2} + 1}}dx}$

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

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

$= \frac{1}{2}\left[ {{x^2}{{\tan }^{ - 1}}x - x + {{\tan }^{ - 1}}x} \right] + c$

$= \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{x}{2} + \frac{1}{2}{\tan ^{ - 1}}x + c$