Evaluate the following integrals:

Let $\mathrm{I}=\int \tan ^{-1}\left(\frac{3 \mathrm{x}-\mathrm{x}^{2}}{1-3 \mathrm{x}^{2}}\right) \mathrm{dx}$

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

We know that, $\frac{3 \tan \theta-\tan \theta^{3}}{1-3 \tan \theta^{2}}=\tan 3 \theta$

$\mathrm{I}=\int \tan ^{-1}\left(\frac{3 \tan \theta-\tan \theta^{3}}{1-3 \tan \theta^{2}}\right) \sec ^{2} \theta \mathrm{d} \theta$

We know that, $\tan ^{-1}(\tan 3 \theta)=3 \theta$

$=\int \tan ^{-1}(\tan 3 \theta) \sec ^{2} \theta d \theta$

$=\int 3 \theta \sec ^{2} \theta d \theta$

Using integration by parts,

$=3\left(\theta \int \sec ^{2} \theta d \theta-\int \frac{d}{d \theta} \theta \int \sec ^{2} \theta d \theta\right)$

$=3\left(\theta \tan \theta-\int \tan \theta d \theta\right)$

$=3(\theta \tan \theta-\log |\sec \theta|)+c$

$=3\left[\operatorname{xtan}^{-1} x+\log \left|\sqrt{1+x^{2}}\right|\right]+c$

$=3 x \tan ^{-1} x+\frac{3}{2} \log \left|1+x^{2}\right|+c$