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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{ dx}$
✓

Answer

Let $\text{I}=\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{ dx}$
Let $\text{x}=\tan\theta$
On differentiating both sides, we get
$\text{dx}=\sec^2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{\sqrt{1+\tan^2\theta}}{\tan^4\theta}\sec^2\theta\text{ d}\theta$
$=\int\frac{\sec^{3}\theta}{\tan^4\theta}\text{ d}\theta$
$=\int\frac{\cos\theta}{\sin^4\theta}\text{ d}\theta$
$=\int\cot\theta\text{cosec}^3\theta\text{ d}\theta$
Let $\text{cosec}^3\theta=\text{t}$
On differentiating both sides, we get
$-3\text{cosec}^3\theta\cot\theta\text{ d}\theta=\text{dt}$
$\therefore\ \text{I}=-\frac{1}{3}\int\cot\theta\text{cosec}^3\theta\frac{\text{dt}}{\text{cosec}^3\theta\cot\theta}$
$=-\frac{\text{t}}{3}+\text{C}$
$=-\frac{1}{3}(\text{cosec}^3\theta)+\text{C}$
$=-\frac{1}{3}(\text{cosec}(\tan^{-1}\text{x}))^3+\text{C}$
$=-\frac{1}{3}\bigg(\text{cosec}\Big(\text{cosec}^1\frac{\sqrt{1+\text{x}^2}}{\text{x}}\Big)\bigg)^3+\text{C}$
$=-\frac{1}{3}\bigg(\frac{\sqrt{1+\text{x}^2}}{\text{x}}\bigg)^3+\text{C}$
Hence, $\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{ dx}=-\frac{1}{3}\bigg(\frac{\sqrt{1+\text{x}^2}}{\text{x}}\bigg)^3+\text{C}$

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