The value of $\int_0^1 \frac{8 \log (1+x)}{1+x^2} d x$ is

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Aicharya · Accepted Answer

$\mathrm{I}=8 \int_0^1 \frac{\log (1+\mathrm{x})}{1+\mathrm{x}^2} \mathrm{dx}$ $=8 \int_0^{\frac{\pi}{4}} \frac{\log (1+	an 	heta)}{1+	an ^2 	heta} \sec ^2 	heta d 	heta(\operatorname{let} x=	an 	heta)$ $=8 \int_0^{\frac{\pi}{4}} \log \left(1+	an \left(\frac{\pi}{4}-	hetaight)ight) d 	heta=8 \int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-	an 	heta}{1+	an 	heta}ight) d 	heta=8 \int_0^{\frac{\pi}{4}} \log 2 d 	heta-8 \int_0^{\frac{\pi}{4}} \log (1+	an 	heta) d 	heta$ $=8 \log 2 \frac{\pi}{4}-I$ $2 \mathrm{I}=2 \pi \log 2$ $\mathrm{I}=\pi \log 2$

The value of $\int_0^1 \frac{8 \log (1+x)}{1+x^2} d x$ is

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