Evaluate the following integrals:

Let $I=\int x^{3} \tan ^{-1} x d x$

Using integration by parts,

We know that,

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

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

$=\tan ^{-1} x \frac{x^{4}}{4}-\frac{1}{4} \int \frac{x^{4}}{1+x^{2}} d x$

$\frac{1}{4} \int \frac{\mathrm{x}^{4}}{1+\mathrm{x}^{2}} \mathrm{dx}=\frac{1}{4}\left[\int \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}+\left(\mathrm{x}^{2}-1\right) \mathrm{dx}\right]=\frac{1}{4}\left[\tan ^{-1} \mathrm{x}+\frac{\mathrm{x}^{3}}{3}-\mathrm{x}\right]$

$=\frac{x^{4}}{4} \tan ^{-1} x-\frac{1}{4}\left[\tan ^{-1} x+\frac{x^{3}}{3}-x\right]+c$