# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \tan ^{-1}(\sqrt{x}) d x$

Solution:

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

$x=t^{2}$

$\mathrm{d} x=2 \mathrm{tdt}$

$I=\int 2 t \tan ^{-1} t d t$

Using integration by parts,

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

We know that,

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

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

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

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

$=\mathrm{t}^{2} \tan ^{-1} \mathrm{t}-\mathrm{t}+\tan ^{-1} \mathrm{t}+\mathrm{c}$

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

$=(\mathrm{x}+1) \tan ^{-1} \sqrt{\mathrm{x}}-\sqrt{\mathrm{x}}+\mathrm{c}$