int frac{x^2-2}{x^3 sqrt{x^2-1}} d x= | vedclass

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(D) Let $I = \int \frac{x^2-2}{x^3 \sqrt{x^2-1}} d x$. Substitute $x = \sec 	heta$,then $dx = \sec 	heta 	an 	heta d	heta$. $\sqrt{x^2-1} = \sqrt

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int \frac{x^2-2}{x^3 \sqrt{x^2-1}} d x$. Substitute $x = \sec 	heta$,then $dx = \sec 	heta 	an 	heta d	heta$. $\sqrt{x^2-1} = \sqrt{\sec^2 	heta - 1} = 	an 	heta$. $I = \int \frac{\sec^2 	heta - 2}{\sec^3 	heta 	an 	heta} \cdot \sec 	heta 	an 	heta d	heta = \int \frac{\sec^2 	heta - 2}{\sec^2 	heta} d	heta = \int (1 - 2 \cos^2 	heta) d	heta$. Using $1 - 2 \cos^2 	heta = -\cos(2	heta)$,we have $I = \int -\cos(2	heta) d	heta = -\frac{1}{2} \sin(2	heta) + C = -\sin 	heta \cos 	heta + C$. Since $\sec 	heta = x$,$\cos 	heta = \frac{1}{x}$ and $\sin 	heta = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2-1}}{x}$. Thus,$I = -\left(\frac{\sqrt{x^2-1}}{x}ight) \left(\frac{1}{x}ight) + C = -\frac{\sqrt{x^2-1}}{x^2} + C$.

$\int \frac{x^2-2}{x^3 \sqrt{x^2-1}} d x=$

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