\(\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=(\) where \(|x| < 1)\)

Let \(I=\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x\)
When \(x=\tan \theta, \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)=\sin ^{-1}(\sin 2 \theta)=2 \theta\) and
\(\mathrm{dx}=\sec ^2 \theta \mathrm{d} \theta \)
\( \therefore \mathrm{I}=\int 2 \theta \sec ^2 \theta \mathrm{d} \theta \)
\( =2 \int \theta \sec ^2 \theta d \theta=2\left[[\theta \tan \theta]-\int \tan \theta d \theta\right] \)
\( =2[\theta \operatorname{tran} \theta+\log |\cos \theta|]+\mathrm{c} \)
\( =2\left(\tan ^{-1} \mathrm{x}\right)(\mathrm{x})+2 \log \left|\sqrt{\frac{1}{1+\tan ^2 \theta}}\right|+\mathrm{c} \)
\( =2 x \tan ^{-1} x+2 \log \left|\sqrt{\frac{1}{1+x^2}}\right|+c \)
\( =2 x \tan ^{-1} x+2 \log \left|1+x^2\right|^{-\frac{1}{2}}+c\) \(=2 x \tan ^{-1} x-\log \left|1+x^2\right|+c\)