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

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(C) Let $I = \int \frac{dx}{(x^2 - 1)\sqrt{x^2 + 1}}$.Put $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$.The integral becomes $\int \frac{\sec

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{dx}{(x^2 - 1)\sqrt{x^2 + 1}}$.Put $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$.The integral becomes $\int \frac{\sec^2 	heta \, d	heta}{(	an^2 	heta - 1)\sec 	heta} = \int \frac{\sec 	heta}{	an^2 	heta - 1} \, d	heta = \int \frac{\cos 	heta}{\sin^2 	heta - \cos^2 	heta} \, d	heta$.Using $\cos^2 	heta = 1 - \sin^2 	heta$,we get $\int \frac{\cos 	heta \, d	heta}{2\sin^2 	heta - 1}$.Let $t = \sin 	heta$,then $dt = \cos 	heta \, d	heta$.$I = \int \frac{dt}{2t^2 - 1} = \frac{1}{2} \int \frac{dt}{t^2 - (1/\sqrt{2})^2}$.Using the formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} ight|$,we get $I = \frac{1}{2} \cdot \frac{1}{2(1/\sqrt{2})} \log \left| \frac{t - 1/\sqrt{2}}{t + 1/\sqrt{2}} ight| + c = \frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{2}t - 1}{\sqrt{2}t + 1} ight| + c$.Since $t = \sin 	heta = \frac{x}{\sqrt{1+x^2}}$,we have $\sqrt{2}t = \frac{x\sqrt{2}}{\sqrt{1+x^2}}$.Substituting this back,$I = \frac{1}{2\sqrt{2}} \log \left| \frac{\frac{x\sqrt{2}}{\sqrt{1+x^2}} - 1}{\frac{x\sqrt{2}}{\sqrt{1+x^2}} + 1} ight| + c = \frac{1}{2\sqrt{2}} \log \left| \frac{x\sqrt{2} - \sqrt{1+x^2}}{x\sqrt{2} + \sqrt{1+x^2}} ight| + c$.Since $\log|a/b| = -\log|b/a|$,this is equivalent to $\frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{1+x^2} - x\sqrt{2}}{\sqrt{1+x^2} + x\sqrt{2}} ight| + c$.

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

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