Let I = int tan^{-1} left( frac{2x}{1-x^2} ri | vedclass

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(B) We know that $	an^{-1} \left( \frac{2x}{1-x^2} ight) = 2 	an^{-1} x$ for $|x| < 1$. Substituting this into the integral,we get $I = \int 2 	a

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$-\log(1+x^2) + c$

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(B) We know that $	an^{-1} \left( \frac{2x}{1-x^2} ight) = 2 	an^{-1} x$ for $|x| Substituting this into the integral,we get $I = \int 2 	an^{-1} x \, dx$. Using integration by parts,let $u = 	an^{-1} x$ and $dv = 2 \, dx$. Then $du = \frac{1}{1+x^2} \, dx$ and $v = 2x$. Applying the formula $\int u \, dv = uv - \int v \, du$: $I = 2x 	an^{-1} x - \int \frac{2x}{1+x^2} \, dx$. Let $t = 1+x^2$,then $dt = 2x \, dx$. So,$\int \frac{2x}{1+x^2} \, dx = \int \frac{1}{t} \, dt = \log|t| + c = \log(1+x^2) + c$. Thus,$I = 2x 	an^{-1} x - \log(1+x^2) + c$. Rearranging the terms,we get $I - 2x 	an^{-1} x = -\log(1+x^2) + c$.

Let $I = \int \tan^{-1} \left( \frac{2x}{1-x^2} \right) dx$,then $I - 2x \tan^{-1} x = $

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