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

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Question

(D) Let $I = \int \sin ^{-1}\left(\frac{2 x}{1+x^2}ight) d x$. Substitute $x = 	an 	heta$,then $dx = \sec ^2 	heta d 	heta$. Since $|x| < 1$,we

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int \sin ^{-1}\left(\frac{2 x}{1+x^2}ight) d x$. Substitute $x = 	an 	heta$,then $dx = \sec ^2 	heta d 	heta$. Since $|x| Thus,$I = \int 2 	heta \sec ^2 	heta d 	heta = 2 \int 	heta \sec ^2 	heta d 	heta$. Using integration by parts,$\int u dv = uv - \int v du$,where $u = 	heta$ and $dv = \sec ^2 	heta d 	heta$: $I = 2 [	heta 	an 	heta - \int 	an 	heta d 	heta] = 2 [	heta 	an 	heta - \log |\sec 	heta|] + c$. Since $	an 	heta = x$ and $\sec 	heta = \sqrt{1+x^2}$,we have $	heta = 	an ^{-1} x$. $I = 2 [x 	an ^{-1} x - \log |\sqrt{1+x^2}|] + c = 2 x 	an ^{-1} x - 2 \log (1+x^2)^{1/2} + c$. $I = 2 x 	an ^{-1} x - \log |1+x^2| + c$.

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

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