Question

Solve:

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

Answer 1

We have,

$I=\int \frac{\sqrt{1+x^2}}{x^4}dx$

Let put

$x=\tan \theta .......\left(1\right)$

Differentiation this and we get,

$dx={\sec }^2\theta d\theta$

Then,

$I=\int \frac{\sqrt{1+{\tan }^2\theta }}{{\tan }^4\theta }{\sec }^2\theta d\theta$

$=\int \frac{\sqrt{{\sec }^2\theta }}{{\tan }^4\theta }{\sec }^2\theta d\theta$

$=\int \frac{{\sec }^3\theta }{{\tan }^4\theta }d\theta$

$=\int \frac{{\sec }^3\theta }{{\tan }^3\theta \tan \theta }d\theta \because \frac{\sec \theta }{\tan \theta }=\csc \theta$

$=\int {\csc }^3\theta \cot \theta d\theta$

$=\int cs{c}^2\theta (\csc \theta \cot \theta )d\theta$

Again let,

$\csc \theta =t$ $......\left(2\right)$

Again differentiating and we get

$-\csc \theta \cot \theta d\theta =dt$

$\csc \theta \cot \theta d\theta =-dt$

Then,

$=\int t^2(-dt)$

$=-\int t^{2}dt$

On integrating and we get,

$=-\frac{t^3}{3}+C\because (\int x^{n}dx=\frac{x^{n+1}}{n+1})$

Put $t=\csc \theta$ by equation (2) to, and we get,

$=-\frac{(\csc \theta {)}^3}{3}+C$

By equation (1) to,

$x=\tan \theta$

$\tan \theta =x$

$\theta ={\tan }^{-1}x$

Then,

$=-\frac{{\left[\csc \left({\tan }^{-1}x\right)\right]}^3}{3}+C$

$=-\frac{{\left[\csc \left({\csc }^{-1}\frac{\sqrt{x^2+1}}{x}\right)\right]}^3}{3}+C\because {\tan }^{-1}x=\frac{\sqrt{x^2+1}}{x}$

$=-\frac{{\left(\frac{\sqrt{x^2+1}}{x}\right)}^3}{3}+C$

$=-\frac{{\left(\sqrt{x^2+1}\right)}^3}{3x^3}+C$

Hence, this is the answer.