Question

Evaluate the following integrals:

$\int \frac{\sqrt{1+x^2}}{x^4}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\sqrt{1+x^2}}{x^4}\mathrm{d}x \\ \mathrm{Let}x=\tan \theta \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \mathrm{d}x={\sec }^2\theta \mathrm{d}\theta \\ \therefore I=\int \frac{\sqrt{1+{\tan }^2\theta }}{{\tan }^4\theta }{\sec }^2\theta \mathrm{d}\theta \\ =\int \frac{{\sec }^3\theta }{{\tan }^4\theta }\mathrm{d}\theta \\ =\int \frac{\cos \theta }{{\sin }^4\theta }\mathrm{d}\theta \\ =\int \cot \theta {\mathrm{cosec}}^3\theta \mathrm{d}\theta \\ \mathrm{Let}{\mathrm{cosec}}^3\theta =t \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ -3{\mathrm{cosec}}^3\theta \cot \theta \mathrm{d}\theta =\mathrm{d}t \\ \therefore I=-\frac{1}{3}\int \cot \theta {\mathrm{cosec}}^3\theta \frac{\mathrm{d}t}{{\mathrm{cosec}}^3\theta \cot \theta } \\ =-\frac{t}{3}+c \\ =-\frac{1}{3}\left({\mathrm{cosec}}^3\theta \right)+c \\ =-\frac{1}{3}{\left(\mathrm{cosec}\left({\tan }^{-1}x\right)\right)}^3+c \\ =-\frac{1}{3}{\left(\mathrm{cosec}\left({\mathrm{cosec}}^{-1}\frac{\sqrt{1+x^2}}{x}\right)\right)}^3+c \\ =-\frac{1}{3}{\left(\frac{\sqrt{1+x^2}}{x}\right)}^3+c \\ \mathrm{Hence},\int \frac{\sqrt{1+x^2}}{x^4}\mathrm{d}x=-\frac{1}{3}{\left(\frac{\sqrt{1+x^2}}{x}\right)}^3+c \end{gathered}$$