Question:
Evaluate the following integrals:
$\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x$
Solution:
let $x=\tan \theta$, so $d x=\sec ^{2} \theta d \theta$ and $\theta=\tan ^{-} x$
Putting above values,
$=\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x=\int \frac{\sqrt{1+\tan ^{2} \theta}}{\tan ^{4} \theta} \sec ^{2} \theta d \theta=\int \sec ^{2} \theta / \tan ^{2} \theta d \theta$
$=\int \operatorname{cosec}^{2} \theta d \theta=-\cot \theta+c$
Put $\theta=\tan ^{-} \mathrm{x}$
$=-\cot \theta+c=-\cot \tan ^{-} x+c$