int frac{1}{x^2 sqrt{1 + x^2}} , dx = | vedclass

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Question

(A) Let $x = 	an 	heta$. Then $dx = \sec^2 	heta \, d	heta$. Substituting these into the integral: $\int \frac{1}{x^2 \sqrt{1 + x^2}} \, dx = \int

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$-\frac{\sqrt{1 + x^2}}{x} + c$

Vedclass · Answer

(A) Let $x = 	an 	heta$. Then $dx = \sec^2 	heta \, d	heta$. Substituting these into the integral: $\int \frac{1}{x^2 \sqrt{1 + x^2}} \, dx = \int \frac{\sec^2 	heta \, d	heta}{	an^2 	heta \sqrt{1 + 	an^2 	heta}} = \int \frac{\sec^2 	heta \, d	heta}{	an^2 	heta \sec 	heta} = \int \frac{\sec 	heta}{	an^2 	heta} \, d	heta$. Simplifying the expression: $\int \frac{1/\cos 	heta}{\sin^2 	heta / \cos^2 	heta} \, d	heta = \int \frac{\cos 	heta}{\sin^2 	heta} \, d	heta = \int \csc 	heta \cot 	heta \, d	heta$. The integral of $\csc 	heta \cot 	heta$ is $-\csc 	heta + c$. Since $x = 	an 	heta$,we have $	an 	heta = \frac{x}{1}$,so $\csc 	heta = \frac{\sqrt{1 + x^2}}{x}$. Thus,the final result is $-\frac{\sqrt{1 + x^2}}{x} + c$.

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

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