Show that

Let $I=\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$

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

When $x=0, \theta=0$ and when $x=1, \theta=\frac{\pi}{4}$

$\begin{aligned} I &=\int_{0}^{\pi} \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right) \sec ^{2} \theta d \theta \\ &=\int_{0}^{\pi} \sin ^{-1}(\sin 2 \theta) \sec ^{2} \theta d \theta \\ &=\int_{0}^{\pi} 2 \theta \cdot \sec ^{2} \theta d \theta \\ &=2 \int_{0}^{\pi} \theta \cdot \sec ^{2} \theta d \theta \end{aligned}$

Taking\thetaas first function and $\sec ^{2} \theta$ as second function and integrating by parts, we obtain

$I=2\left[\theta \int \sec ^{2} \theta d \theta-\int\left\{\left(\frac{d}{d x} \theta\right) \int \sec ^{2} \theta d \theta\right\} d \theta\right]_{0}^{\frac{\pi}{4}}$

$=2\left[\theta \tan \theta-\int \tan \theta d \theta\right]_{0}^{\frac{\pi}{4}}$

$=2[\theta \tan \theta+\log |\cos \theta|]_{0}^{\frac{\pi}{4}}$

$=2\left[\frac{\pi}{4} \tan \frac{\pi}{4}+\log \left|\cos \frac{\pi}{4}\right|-\log |\cos 0|\right]$

$=2\left[\frac{\pi}{4}+\log \left(\frac{1}{\sqrt{2}}\right)-\log 1\right]$

$=2\left[\frac{\pi}{4}-\frac{1}{2} \log 2\right]$

$=\frac{\pi}{2}-\log 2$