Evaluate the following integrals:

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

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

$\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)=\sin ^{-1}(\sin 2 \theta)=2 \theta$

$\int \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x=\int 2 \theta \sec ^{2} \theta d \theta$

Using integration by parts,

$=2\left(\theta \int \sec ^{2} \theta d \theta-\int \frac{d}{d \theta} \theta \int \sec ^{2} \theta d \theta\right)$

$=2\left(\theta \tan \theta-\int \tan \theta d \theta\right)$

We know that, $\int \tan \theta d \theta=\log |\cos \theta|$

$=2(\theta \tan \theta-\log |\cos \theta|)+\mathrm{c}$\

$=2\left[x \tan ^{-1} x+\log \left|\frac{1}{\sqrt{1+x^{2}}}\right|\right]+c$

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

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

$=2 x \tan ^{-1} x+\log (1+x)^{2}+c$