\(I=\int_0^1 \frac{8 \log (1+x)}{1+x^2} d x=8 \int_0^1 \frac{\log (1+x)}{1+x^2} \cdot d x\)
Let \(x=\tan \theta d x=\sec ^2 \theta d \theta\)
When \(x=0, \theta=0\)
and when \(x=1, \theta=\frac{\pi}{4}\)
So,
\(\begin{aligned}
I & =8 \int_0^{\pi / 4} \frac{\log (1+\tan \theta)}{1+\tan ^2 \theta} \cdot \sec ^2 \theta \cdot d \theta \\
& =8 \int_0^\pi \log (1+\tan \theta) d \theta
\end{aligned}\)
Now using
\(I=\int_0^a f(x) d x=\int_0^a f(a-x) d x\)
and adding both equations we have,
\(\begin{aligned}
2 I & =8 \int_0^\pi \log 2 d x \\
I & =8 \times \frac{\pi}{8} \log 2=\pi \log 2
\end{aligned}\)