Question

Evaluate the following integrals:

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

Answer 1

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

$$\text{Let x}=\tan \theta$$

$$\text{On differentiating both sides, we get}$$

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

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

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

$$=\int \frac{\cos \theta }{{\sin }^4\theta }d\theta$$

${\displaystyle =\int \cot \theta {\text{cosec}}^3\theta d\theta }$

${\displaystyle Let {\text{cosec}}^3\theta =t}$ 

$$\text{On differentiating both sides, we get}$$

${\displaystyle -3{\text{cosec}}^3\theta \cot \theta d\theta =dt}$

$$\therefore I=-\frac{1}{3}\int \cot \theta {\text{ cosec }}^3\theta \frac{dt}{{cosec}^3\theta \cot \theta }$$

$$=-\frac{t}{3}+c$$

$$=-\frac{1}{3}\left({cosec}^3\theta \right)+c$$

$$=-\frac{1}{3}{\left(cosec({\tan }^{-1}x)\right)}^3+c$$

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

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

$$Hence,\int \frac{\sqrt{1+x^2}}{x^4}dx=-\frac{1}{3}{\left(\frac{\sqrt{1+x^2}}{x}\right)}^3+c$$