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

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

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$2 x 	an ^{-1} x$

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(A) Let $I = \int \sin ^{-1}\left(\frac{2 x}{1+x^2}ight) d x$.Substitute $x = 	an 	heta$, then $d x = \sec ^2 	heta d 	heta$.Since $\sin ^{-1}\left(\frac{2 	an 	heta}{1+	an ^2 	heta}ight) = \sin ^{-1}(\sin 2 	heta) = 2 	heta$ (assuming $|x| \le 1$), we have:$I = \int 2 	heta \sec ^2 	heta d 	heta$.Using integration by parts, $\int u dv = uv - \int v du$, where $u = 	heta$ and $dv = \sec ^2 	heta d 	heta$:$I = 2 \left[ 	heta 	an 	heta - \int 	an 	heta d 	heta ight]$.$I = 2 [	heta 	an 	heta + \log |\cos 	heta|] + C$.Since $	an 	heta = x$, we have $	heta = 	an ^{-1} x$ and $\cos 	heta = \frac{1}{\sqrt{1+x^2}}$.$I = 2 [x 	an ^{-1} x + \log |\frac{1}{\sqrt{1+x^2}}|] + C$.$I = 2 x 	an ^{-1} x + 2 \log (1+x^2)^{-1/2} + C$.$I = 2 x 	an ^{-1} x - \log (1+x^2) + C$.Comparing this with $f(x) - \log (1+x^2)$, we get $f(x) = 2 x 	an ^{-1} x$.

If $\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=f(x)-\log \left(1+x^2\right)$, then $f(x)$ is equal to

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