Let $I=\int_{0}^{1} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) d x$
Put $x=\tan \theta \Rightarrow d x=\sec ^{2} \theta d \theta$
When $x=0, \theta=0$ and when $x=1, \theta=\frac{\pi}{4}$
$\therefore \mathrm{I}=\int_{0}^{\frac{\pi}{4}}\left[\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)\right]\left(1+\tan ^{2} \theta\right) \mathrm{d} \theta=\int_{0}^{\frac{\pi}{4}} \tan ^{-1}(\tan 2 \theta)\left(1+\tan ^{2} \theta\right) \mathrm{d} \theta$
$=\int_{0}^{\frac{\pi}{4}} 2 \theta \sec ^{2} \theta d \theta=2 \int_{0}^{\pi / 4} \theta \sec ^{2} \theta d \theta=2[\theta \tan \theta]_{0}^{\pi / 4}-2 \int_{0}^{\pi / 4} \tan \theta d \theta$
$=2\left[\frac{\pi}{4}\right]+2[\log |\cos \theta|]_{0}^{\pi / 4}=\frac{\pi}{2}+2\left[\log \left|\frac{1}{\sqrt{2}}\right|\right]$
$=\frac{\pi}{2}+2 \log (2)^{\frac{1}{2}}$
$=\frac{\pi}{2}-\log 2$