Show that

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

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

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

Integrating by parts, we obtain

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

$=2\left[\theta \cdot \tan \theta-\int \tan \theta d \theta\right]$

$=2[\theta \tan \theta+\log |\cos \theta|]+\mathrm{C}$

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

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

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

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