Question

Evaluate the following integrals:

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

Answer 1

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

\[ \text{Let x} = \tan\theta\]

\[ \text{On differentiating both sides, we get}\]

` dx = sec^2  θ   dθ `

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

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

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

` = ∫ cot  θ   "cosec"^3 θ  dθ `

` Let   "cosec"^3θ = t` 

\[ \text{On differentiating both sides, we get}\]

`  - 3 \text{ cosec}^3 θ    cot θ  dθ = dt `

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

\[ = - \frac{t}{3} + c\]

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

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

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

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

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