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,
$=\tan ^{-1} x \int x^{2} d x-\int\left(\frac{1}{1+x^{2}}\right) \int x^{2} d x$
$=\tan ^{-1} x \frac{x^{3}}{3}-\frac{1}{3} \int \frac{x^{3}}{1+x^{2}} d x$
$=\tan ^{-1} x \frac{x^{3}}{3}-\frac{1}{3} \int x-\frac{x}{1+x^{2}} d x$
$=\tan ^{-1} x \frac{x^{3}}{3}-\frac{1}{3} \times \frac{x^{2}}{2}+\int \frac{x}{1+x^{2}} d x$
$=\frac{1}{3} x^{3} \tan ^{-1} x-\frac{x^{2}}{6}+\frac{1}{6} \log \left|1+x^{2}\right|+c$