Evaluate the following integrals:

$\int x^{2} \tan ^{-1} x d x$

Here we will use integration by parts,

$\int u \cdot d v=u v-\int v d u$

Choose $u$ in these oder LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG,E-EXPONENTIAL)

So here, $u=\tan ^{-1} x$

$=\tan ^{-1} x \int x^{2} d x-\frac{1}{3} \int x^{3}\left(d\left(\tan ^{-1} x\right)\right) /$

$d x+c$

$\left.\int x^{2} d x=\left(\frac{x^{3}}{3}\right)+c\right)$

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

Putting $1+x^{2}=t$,

$2 x d x=d t$

$x d x=\frac{d t}{2}$

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

$=\left(\frac{x^{3}}{3}\right) \tan ^{-1} x-\frac{1}{3} \int \frac{(t-1)}{t} \frac{d t}{2}$

$=\left(\frac{x^{3}}{3}\right) \tan ^{-1} x-\frac{1}{6} \int \frac{(t-1)}{t} d t$

$=\left(\frac{x^{3}}{3}\right) \tan ^{-1} x-\frac{1}{6}\left[\int 1 d t-\int \frac{1}{t} d t\right]$

$=\left(\frac{x^{3}}{3}\right) \tan ^{-1} x-\frac{1}{6}[-\log t+t]+c$

Resubstituting $\mathrm{t}$

$=\left(\frac{x^{3}}{3}\right) \tan ^{-1} x-\frac{1}{6}\left[-\log \left(1+x^{2}\right)+\left(1+x^{2}\right)\right]+c$