The value of ${\int }_0^{1}8\frac{\ln (1+x)}{1+x^2}dx$ is:

Let $x=\tan \theta$, then $dx={\sec }^2\theta d\theta$.Limits: $x=0\to \theta =0$; $x=1\to \theta =\pi /4$.
$$I=8{\int }_0^{\pi /4}\frac{\ln (1+\tan \theta )}{1+{\tan }^2\theta }{\sec }^2\theta d\theta =8{\int }_0^{\pi /4}\ln (1+\tan \theta )d\theta$$
Using property ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$:
$$I=8{\int }_0^{\pi /4}\ln [1+\tan (\pi /4-\theta )]d\theta =8{\int }_0^{\pi /4}\ln \left[1+\frac{1-\tan \theta }{1+\tan \theta }\right]d\theta$$
$$I=8{\int }_0^{\pi /4}\ln \left[\frac{2}{1+\tan \theta }\right]d\theta =8{\int }_0^{\pi /4}[\ln 2-\ln (1+\tan \theta )]d\theta$$
$$I=8(\ln 2)(\pi /4)-I⟹2I=2\pi \ln 2⟹I=\pi \ln 2$$
Check source solution: The source indicates $I=\pi \ln 2$ (after integrating $8\ln 2$).