Evaluate the following integrals:

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

$x^{2}=t$

$2 x d x=d t$

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

Using integration by parts,

$=\frac{1}{2}\left(\tan ^{-1} \mathrm{t} \int \mathrm{dt}-\int \frac{\mathrm{d}}{\mathrm{dt}} \tan ^{-1} \mathrm{t} \int \mathrm{dt}\right)$

We know that,

$\frac{d}{d t} \tan ^{-1} t=\frac{1}{2\left(1+t^{2}\right)}$

$=\frac{1}{2}\left[\mathrm{t} \tan ^{-1} \mathrm{t}-\int \frac{\mathrm{t}}{\left(1+\mathrm{t}^{2}\right)} \mathrm{dt}\right]$

$=\frac{\mathrm{t}}{2} \tan ^{-1} \mathrm{t}-\frac{1}{4} \int \frac{2 \mathrm{t}}{1+\mathrm{t}^{2}} \mathrm{dt}$

$=\frac{\mathrm{t}}{2} \tan ^{-1} \mathrm{t}-\frac{1}{4} \log \left|1+\mathrm{t}^{2}\right|+\mathrm{c}$

$=\frac{\mathrm{x}^{2}}{2} \tan ^{-1} \mathrm{x}^{2}-\frac{1}{4} \log \left|1+\mathrm{x}^{4}\right|+\mathrm{c}$