7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

$\because \,\,\,x{{\tan }^{-1}}x$  is an even function

$I = [2\frac{{{x^2}}}{2}{\tan ^{ - 1}}x]_0^1 - 2\int_0^1 {\frac{1}{2}\frac{{{x^2}}}{{1 + {x^2}}}dx} $

$I = [{x^2}{\tan ^{ - 1}}x]_0^1 - \int_0^1 {\frac{{{x^2} + 1 - 1}}{{1 + {x^2}}}dx} $

$I = [{x^2}{\tan ^{ - 1}}x]_0^1 - [x]_0^1 + [{\tan ^{ - 1}}x]_0^1$

==> $I = \frac{\pi }{4} - 1 + \frac{\pi }{4} = \frac{\pi }{2} - 1$.