int frac{1}{left(1+x^2right) sqrt{x^2+2}} d x | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{1}{(1+x^2) \sqrt{x^2+2}} dx$.Substitute $x = \sqrt{2} 	an 	heta$,then $dx = \sqrt{2} \sec^2 	heta d	heta$.$I = \int \frac{\sqrt{2} \sec^2 	heta}{(1 + 2 	an^2 	heta) \sqrt{2 	an^2 	heta + 2}} d	heta = \int \frac{\sqrt{2} \sec^2 	heta}{(1 + 2 	an^2 	heta) \sqrt{2} \sec 	heta} d	heta = \int \frac{\sec 	heta}{1 + 2 	an^2 	heta} d	heta$.Using $1 + 2 	an^2 	heta = 1 + 2(\sec^2 	heta - 1) = 2 \sec^2 	heta - 1$,we get $I = \int \frac{\sec 	heta}{2 \sec^2 	heta - 1} d	heta = \int \frac{\cos 	heta}{2 - \cos^2 	heta} d	heta = \int \frac{\cos 	heta}{1 + \sin^2 	heta} d	heta$.Let $t = \sin 	heta$,then $dt = \cos 	heta d	heta$.$I = \int \frac{1}{1 + t^2} dt = 	an^{-1}(t) + c = 	an^{-1}(\sin 	heta) + c$.Since $x = \sqrt{2} 	an 	heta$,$	an 	heta = \frac{x}{\sqrt{2}}$. Using a triangle,$\sin 	heta = \frac{x}{\sqrt{x^2+2}}$.Thus,$I = 	an^{-1} \left( \frac{x}{\sqrt{x^2+2}} ight) + c$.Note: The provided options suggest a different form. Re-evaluating the substitution $x = \frac{1}{t}$ or similar leads to the form $-	an^{-1} \frac{\sqrt{x^2+2}}{|x|} + c$.

$\int \frac{1}{\left(1+x^2\right) \sqrt{x^2+2}} d x=$

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