# Evaluate the following integrals:

Question:

Evaluate the following integrals:

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

Solution:

Let $\mathrm{I}=\int \mathrm{x}^{2} \tan ^{-1} \mathrm{x} \mathrm{dx}$

Using integration by parts,

Taking inverse function as first function and algebraic function as second function,

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

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

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

$=\tan ^{-1} x \frac{x^{3}}{3}-\frac{1}{3} \times \frac{x^{2}}{2}+\int \frac{x}{1+x^{2}} d x$

$=\frac{1}{3} x^{3} \tan ^{-1} x-\frac{x^{2}}{6}+\frac{1}{6} \log \left|1+x^{2}\right|+c$