$\int\limits_0^1 {{{\sin }^{ - 1}}\left( {\cfrac{{2x}}{{1 + {x^2}}}} \right)dx}$

Let $I = \int\limits_0^1 {{{\sin }^{ - 1}}\left( {\cfrac{{2x}}{{1 + {x^2}}}} \right)dx}$

$= \int\limits_0^{\pi /2} {\sqrt {\sin \phi } {{\left( {1 - {{\sin }^2}\phi } \right)}^2}\cos \phi } d\phi$

Put $x = \tan \theta$ $\Rightarrow$ $dx = {\sec ^2}\theta d\theta$

When $x = 0,\theta = 0$ and when $x = 1,\theta = \cfrac{\pi }{4}$
$\therefore$

$I = \int\limits_0^{\pi /4} {{{\sin }^{ - 1}}\left( {\cfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)} {\sec ^2}\theta d\theta$

$= \int\limits_0^{\pi /4} {{{\sin }^{ - 1}}\left( {\sin 2\theta } \right).} {\sec ^2}\theta d\theta = \int\limits_0^{\pi /4} {2\theta {{\sec }^2}\theta d\theta }$

$= \left[ {2\theta \tan \theta - 2\int {\left( {\cfrac{d}{{d\theta }}\left( \theta \right) \cdot \tan \theta } \right)} d\theta } \right]_0^{\pi /4} = \left[ {2\theta \tan \theta - 2\log \sec \theta } \right]_0^{\pi /4}$

$= \left( {2 \cdot \cfrac{\pi }{4} \cdot \tan \cfrac{\pi }{4} - 2\log \sec \cfrac{\pi }{4}} \right) - \left( {0 - 2\log 1} \right)$

$= \cfrac{\pi }{2} - 2\log \sqrt 2 = \cfrac{\pi }{2} - \log 2$