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

\(\begin{aligned} & \text {let } x=\tan \theta \\ & \Rightarrow \mathrm{d} x=\sec ^2 \theta \mathrm{d} \theta \text {, when } x=0, \theta=0 \text {; when } x=1 \text {, } \\ & \theta=\frac{\pi}{4}\end{aligned}\)
\(I=\int_0^{\frac{\pi}{4}} 8 \frac{\log (1+\tan \theta)}{1+\tan ^2 \theta} \cdot \sec ^2 \theta d \theta=\int_0^{\frac{\pi}{4}} 8 \log (1+\tan \theta) \mathrm{d} \theta \quad\) \(\ldots \ldots(\mathrm{i})\)
\(\Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{4}} 8 \log \left(1+\tan(\frac{\pi}{4}-\theta\right) \mathrm{d} \theta[\because \int_0^a f(x) \mathrm{d} x=\) \(\int_0^a f(a-x) \mathrm{d} x]\)

\(\left[\because \tan \left(\frac{\pi}{4}-\theta\right)=\frac{1-\tan \theta}{1+\tan \theta}\right]\)
from (i) \(+(\) ii \()\)
\(\begin{aligned}
& 2 I=\int_0^{\pi / 4} 8 \log \left\{(1+\tan \theta) \times \frac{2}{(1+\tan \theta)}\right\} d \theta \\
& \Rightarrow 2 \mathrm{I}=8 \int_0^{\pi / 4} \log 2 \mathrm{~d} \theta=8 \log 2[\theta]_0^{\pi / 4}=8 \log 2 \times \frac{\pi}{4}
\end{aligned}\)
\(\Rightarrow \mathrm{I}=\pi \log 2\)