Evaluate the following integrals:

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

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

$I=\int \tan ^{-1}\left(\frac{2 \tan \theta}{1-2 \tan \theta^{2}}\right) \sec ^{2} \theta d \theta$

We know that, $\frac{2 \tan \theta}{1-2 \tan \theta^{2}}=\tan 2 \theta$

$=\int \tan ^{-1}(\tan 2 \theta) \sec ^{2} \theta d \theta$

$\int 2 \theta \sec ^{2} \theta d \theta$

Using integration by parts,

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

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

$=2(\theta \tan \theta-\log |\sec \theta|)+c$

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

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