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प्रश्न
Evaluate the following integrals:
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उत्तर
\[\text{ Let I } = \int\frac{\sqrt{1 + x^2}}{x^4}dx\]
\[ \text{Let x} = \tan\theta\]
\[ \text{On differentiating both sides, we get}\]
` 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 `
\[ \therefore I = - \frac{1}{3}\int\cotθ \text{ cosec }^3 \theta \frac{dt}{{cosec}^3 \theta \cot\theta}\]
\[ = - \frac{t}{3} + c\]
\[ = - \frac{1}{3}\left( {cosec}^3 \theta \right) + c\]
\[ = - \frac{1}{3} \left( cosec\left( \tan^{- 1} x \right) \right)^3 + c\]
\[ = - \frac{1}{3} \left( cosec\left( {cosec}^{- 1} \frac{\sqrt{1 + x^2}}{x} \right) \right)^3 + c\]
\[ = - \frac{1}{3} \left( \frac{\sqrt{1 + x^2}}{x} \right)^3 + c\]
\[Hence, \int\frac{\sqrt{1 + x^2}}{x^4}dx = - \frac{1}{3} \left( \frac{\sqrt{1 + x^2}}{x} \right)^3 + c\]
