$\int_0^1 \frac{8 \log (1+x)}{1+x^2} \mathrm{~d} x=$

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { let } x=	an 	heta \ & \Rightarrow \mathrm{d} x=\sec ^2 	heta \mathrm{d} 	heta 	ext {, when } x=0, 	heta=0 	ext {; when } x=1 	ext {, } \ & 	heta=\frac{\pi}{4}\end{aligned}$ $I=\int_0^{\frac{\pi}{4}} 8 \frac{\log (1+	an 	heta)}{1+	an ^2 	heta} \cdot \sec ^2 	heta d 	heta=\int_0^{\frac{\pi}{4}} 8 \log (1+	an 	heta) \mathrm{d} 	heta \quad \ldots \ldots(\mathrm{i})$ $\Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{4}} 8 \log \left(1+	an \left(\frac{\pi}{4}-	hetaight) \mathrm{d} 	heta\left[\because \int_0^a f(x) \mathrm{d} x=\int_0^a f(a-x) \mathrm{d} xight]ight.$ $\left[\because 	an \left(\frac{\pi}{4}-	hetaight)=\frac{1-	an 	heta}{1+	an 	heta}ight]$ from (i) $+($ ii $)$ $\begin{aligned} & 2 I=\int_0^{\pi / 4} 8 \log \left\{(1+	an 	heta) 	imes \frac{2}{(1+	an 	heta)}ight\} d 	heta \ & \Rightarrow 2 \mathrm{I}=8 \int_0^{\pi / 4} \log 2 \mathrm{~d} 	heta=8 \log 2[	heta]_0^{\pi / 4}=8 \log 2 	imes \frac{\pi}{4} \end{aligned}$ $\Rightarrow \mathrm{I}=\pi \log 2$

$\int_0^1 \frac{8 \log (1+x)}{1+x^2} \mathrm{~d} x=$

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