$\int_{0}^{1} an ^{-1}\left(\frac{2 x}{1-x^{2}} ight) d x=$

Question

$\int_{0}^{1} 	an ^{-1}\left(\frac{2 x}{1-x^{2}}ight) d x=$

Aicharya · Accepted Answer

Let $I=\int_{0}^{1} 	an ^{-1}\left(\frac{2 x}{1-x^{2}}ight) d x$ Put $x=	an 	heta \Rightarrow d x=\sec ^{2} 	heta d 	heta$ When $x=0, 	heta=0$ and when $x=1, 	heta=\frac{\pi}{4}$ $	herefore \mathrm{I}=\int_{0}^{\frac{\pi}{4}}\left[	an ^{-1}\left(\frac{2 	an 	heta}{1-	an ^{2} 	heta}ight)ight]\left(1+	an ^{2} 	hetaight) \mathrm{d} 	heta=\int_{0}^{\frac{\pi}{4}} 	an ^{-1}(	an 2 	heta)\left(1+	an ^{2} 	hetaight) \mathrm{d} 	heta$ $=\int_{0}^{\frac{\pi}{4}} 2 	heta \sec ^{2} 	heta d 	heta=2 \int_{0}^{\pi / 4} 	heta \sec ^{2} 	heta d 	heta=2[	heta 	an 	heta]_{0}^{\pi / 4}-2 \int_{0}^{\pi / 4} 	an 	heta d 	heta$ $=2\left[\frac{\pi}{4}ight]+2[\log |\cos 	heta|]_{0}^{\pi / 4}=\frac{\pi}{2}+2\left[\log \left|\frac{1}{\sqrt{2}}ight|ight]$ $=\frac{\pi}{2}+2 \log (2)^{\frac{1}{2}}$ $=\frac{\pi}{2}-\log 2$

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

Solution