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

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(D) Let $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$. Substituting these into the integral: $\int 	an^{-1} \left( \frac{2 	an 	heta}{1 -

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$. Substituting these into the integral: $\int 	an^{-1} \left( \frac{2 	an 	heta}{1 - 	an^2 	heta} ight) \sec^2 	heta \, d	heta = \int 	an^{-1}(	an 2	heta) \sec^2 	heta \, d	heta$ $= \int 2	heta \sec^2 	heta \, d	heta = 2 \int 	heta \sec^2 	heta \, d	heta$. Using integration by parts: $= 2 \left[ 	heta 	an 	heta - \int 	an 	heta \, d	heta ight] = 2 \left[ 	heta 	an 	heta - \log|\sec 	heta| ight] + c$ $= 2 	heta 	an 	heta - 2 \log|\sec 	heta| + c$. Since $	an 	heta = x$,then $	heta = 	an^{-1} x$ and $\sec 	heta = \sqrt{1 + x^2}$. $= 2x 	an^{-1} x - 2 \log(\sqrt{1 + x^2}) + c$ $= 2x 	an^{-1} x - 2 \cdot \frac{1}{2} \log(1 + x^2) + c$ $= 2x 	an^{-1} x - \log(1 + x^2) + c$.

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

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