int frac{dx}{x^2 sqrt{4+x^2}} is equal to | vedclass

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(C) Let $I = \int \frac{dx}{x^2 \sqrt{4+x^2}}$.Substitute $x = 2 	an 	heta$,then $dx = 2 \sec^2 	heta \ d	heta$.The integral becomes $I = \int \fr

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$\frac{-1}{4x} \sqrt{4+x^2}+C$

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(C) Let $I = \int \frac{dx}{x^2 \sqrt{4+x^2}}$.Substitute $x = 2 	an 	heta$,then $dx = 2 \sec^2 	heta \ d	heta$.The integral becomes $I = \int \frac{2 \sec^2 	heta \ d	heta}{(4 	an^2 	heta) \sqrt{4 + 4 	an^2 	heta}}$.Since $\sqrt{4(1 + 	an^2 	heta)} = 2 \sec 	heta$,we have $I = \int \frac{2 \sec^2 	heta \ d	heta}{4 	an^2 	heta \cdot 2 \sec 	heta} = \frac{1}{4} \int \frac{\sec 	heta}{	an^2 	heta} \ d	heta$.$I = \frac{1}{4} \int \frac{1/\cos 	heta}{\sin^2 	heta / \cos^2 	heta} \ d	heta = \frac{1}{4} \int \frac{\cos 	heta}{\sin^2 	heta} \ d	heta$.Let $u = \sin 	heta$,then $du = \cos 	heta \ d	heta$. So $I = \frac{1}{4} \int u^{-2} \ du = \frac{1}{4} (-u^{-1}) + C = -\frac{1}{4 \sin 	heta} + C$.Since $	an 	heta = \frac{x}{2}$,we have $\sin 	heta = \frac{x}{\sqrt{x^2+4}}$.Thus,$I = -\frac{1}{4} \cdot \frac{\sqrt{x^2+4}}{x} + C = -\frac{\sqrt{4+x^2}}{4x} + C$.

$\int \frac{dx}{x^2 \sqrt{4+x^2}}$ is equal to

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