Put $x=\tan t$ so that $d x=\sec ^2 t d t$
$\begin{aligned}
&\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=\int \sin ^{-1}\left(\frac{2 \tan \mathrm{t}}{1+\tan ^2 \mathrm{t}}\right) \sec ^2 \mathrm{t} d \mathrm{t} \\
&=\int \sin ^{-1}(\sin 2 \mathrm{t}) \sec ^2 \mathrm{t} d \mathrm{t}=\int 2 \mathrm{t} \sec ^2 \mathrm{t} \mathrm{dt} \\
&=2\left[\mathrm{t} \tan \mathrm{t}-\int 1 \cdot \tan \mathrm{t} \mathrm{dt}\right] \\
&=2 \mathrm{t} \tan \mathrm{t}+2 \log |\cos \mathrm{t}|+\mathrm{c} \\
&=2 \mathrm{x} \tan ^{-1} \mathrm{x}-\log \left(1+\mathrm{x}^2\right)+\mathrm{c}
\end{aligned}$