On I subset R-{-1,1}, int tan ^{-1}left(frac{ | vedclass

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(C) Let $I_1 = \int 	an ^{-1}\left(\frac{2 x}{1-x^2}ight) d x$. For $|x| < 1$, we know that $	an ^{-1}\left(\frac{2 x}{1-x^2}ight) = 2 	an ^{-1

Vedclass · Accepted Answer

$x 	an ^{-1}\left(\frac{2 x}{1-x^2}ight)-\log \left(1+x^2ight)+c$

Vedclass · Answer

(C) Let $I_1 = \int 	an ^{-1}\left(\frac{2 x}{1-x^2}ight) d x$. For $|x| Thus, $I_1 = \int 2 	an ^{-1} x d x = 2 \int 	an ^{-1} x d x$. Using integration by parts, let $u = 	an ^{-1} x$ and $dv = dx$. Then $du = \frac{1}{1+x^2} dx$ and $v = x$. $I_1 = 2 \left( x 	an ^{-1} x - \int \frac{x}{1+x^2} dx ight)$. $I_1 = 2 x 	an ^{-1} x - \int \frac{2x}{1+x^2} dx$. $I_1 = 2 x 	an ^{-1} x - \log(1+x^2) + C$. Since $2 	an ^{-1} x = 	an ^{-1}\left(\frac{2 x}{1-x^2}ight)$, we substitute back: $I_1 = x 	an ^{-1}\left(\frac{2 x}{1-x^2}ight) - \log(1+x^2) + C$.

On $I \subset R-\{-1,1\}, \int \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) d x=$

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