$\underset{0}{\overset{1}{\int }}\frac{8\log (1+x)}{1+x^2}dx=$

$\underset{0}{\overset{1}{\int }}\frac{8\log (1+x)}{1+x^2}dx$

Put $x=\tan \theta$

$\Rightarrow \theta ={\tan }^{-1}x$

$\Rightarrow d\theta =\frac{1}{1+x^2}dx$

$x=0\Rightarrow \tan \theta =0\Rightarrow \theta =0$

$x=1\Rightarrow \tan \frac{\pi }{4}=1\Rightarrow \theta =\frac{\pi }{4}$

$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{8\log (1+\tan \theta )}{1+x^2}\cdot (1+x^2)\cdot d\theta \to \left(1\right)$

$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}8\log [1+\tan (\frac{\pi }{4}-\theta )]d\theta \to \left(2\right)$            $(\because King\text{'}srule)$

$\left(1\right)+\left(2\right)$

$\Rightarrow 2I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}8\log 2d\theta$

$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}4\log 2d\theta$

$=4\log 2{\left(\theta \right)}_0^{\pi /4}$

$=4\log 2(\frac{\pi }{4}-0)$

$=\pi \log 2$.