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

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(A) Let $I = \int \sin ^{-1}\left(\frac{2 x}{1+x^2}ight) d x$.Substitute $x = 	an t$,then $dx = \sec^2 t dt$.The integral becomes $I = \int \sin^{-

Vedclass · Accepted Answer

$2 x 	an ^{-1} x-\log \left(1+x^2ight)+c$,where $c$ is a constant of integration.

Vedclass · Answer

(A) Let $I = \int \sin ^{-1}\left(\frac{2 x}{1+x^2}ight) d x$.Substitute $x = 	an t$,then $dx = \sec^2 t dt$.The integral becomes $I = \int \sin^{-1}(\sin 2t) \sec^2 t dt = \int 2t \sec^2 t dt$.Using integration by parts: $\int u dv = uv - \int v du$,where $u = 2t$ and $dv = \sec^2 t dt$.$I = 2t 	an t - \int 2 	an t dt = 2t 	an t - 2 \log |\sec t| + c$.Since $\sec t = \sqrt{1 + 	an^2 t} = \sqrt{1 + x^2}$,we have $\log |\sec t| = \log (1 + x^2)^{1/2} = \frac{1}{2} \log (1 + x^2)$.Substituting back: $I = 2x 	an^{-1} x - 2(\frac{1}{2} \log (1 + x^2)) + c = 2x 	an^{-1} x - \log (1 + x^2) + c$.

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

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