int_{0}^{1} tan^{-1}left(frac{2x}{1-x^2}right | vedclass

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(B) Let $I = \int_{0}^{1} 	an^{-1}\left(\frac{2x}{1-x^2}ight) dx$.Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta d	heta$.When $x = 0$,$	he

Vedclass · Accepted Answer

$\frac{\pi}{2} - \log 2$

Vedclass · Answer

(B) Let $I = \int_{0}^{1} 	an^{-1}\left(\frac{2x}{1-x^2}ight) dx$.Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta d	heta$.When $x = 0$,$	heta = 0$; when $x = 1$,$	heta = \frac{\pi}{4}$.The integral becomes $I = \int_{0}^{\pi/4} 	an^{-1}(	an 2	heta) \sec^2 	heta d	heta = \int_{0}^{\pi/4} 2	heta \sec^2 	heta d	heta$.Using integration by parts: $I = 2 \left[ 	heta 	an 	heta - \int 	an 	heta d	heta ight]_{0}^{\pi/4}$.$I = 2 \left[ 	heta 	an 	heta + \log |\cos 	heta| ight]_{0}^{\pi/4}$.$I = 2 \left[ (\frac{\pi}{4} \cdot 1 + \log |\frac{1}{\sqrt{2}}|) - (0 + \log 1) ight]$.$I = 2 \left[ \frac{\pi}{4} - \frac{1}{2} \log 2 ight] = \frac{\pi}{2} - \log 2$.

$\int_{0}^{1} \tan^{-1}\left(\frac{2x}{1-x^2}\right) dx =$

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