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

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

Put $x = \tan \,t$ $\Rightarrow$ $dx = {\sec ^2}t\,dt$

$\therefore$ $I = \int {{{\sin }^{ - 1}}\left( {\cfrac{{2\tan \,t}}{{1 + {{\tan }^2}t}}} \right){{\sec }^2}t} dt = \int {{{\sin }^{ - 1}}\left( {\sin 2t} \right){{\sec }^2}t} dt$

$= \int {2t{{\sec }^2}t\,dt} = 2\int {t\,{{\sec }^2}t\,dt}$

$= 2\left[ {t\int {{{\sec }^2}t\,dt - } \int {\left( {\cfrac{d}{{dt}}\left( t \right) \cdot \int {{{\sec }^2}t\,dt} } \right)dt} } \right]$

$= 2\left[ {t\tan t - \int {1 \cdot \tan \,t\,dt} } \right]$

$= 2t\tan t + 2\log \left| {\cos t} \right| + C = 2{\tan ^{ - 1}}x \cdot x + 2\log \left| {\cfrac{1}{{\sqrt {1 + {x^2}} }}} \right| + C$

$= 2x{\tan ^{ - 1}}x + 2\log \left[ {{{\left( {1 + {x^2}} \right)}^{\cfrac{1}{2}}}} \right] + C$

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

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

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