The integral ${\int }_0^{1/a} \frac{\log (1+ax)}{1+a^2{x}^2}\mathrm{d}x(a>0)$ is equal to 

Put $ax=t,$ the given integral reduces to

$\begin{array}{l}\frac{1}{a}{\int }_0^1 \frac{\log (1+t)}{1+t^2}dt=\frac{1}{a}{\int }_0^{\pi /4} \frac{\log (1+\tan u)}{1+{\tan }^2 u}{\sec }^2 udu(t-\tan u)\\ =\frac{1}{a}{\int }_0^{\pi /4} \log (1+\tan u)du\\ =\frac{1}{a}{\int }_0^{\pi /4} \log \left(1+\tan \left(\frac{\pi }{4}-u\right)\right)du\\ =\frac{1}{a}{\int }_0^{\pi /4} \log \left(1+\frac{1-\tan u}{1+\tan u}\right)du\\ =\frac{1}{a}{\int }_0^{\pi /4} [\log 2-\log (1+\tan u\left)\right]du\\ \Rightarrow \frac{2}{a}{\int }_0^{\pi /4} \log (1+\tan u)du=\frac{1}{a}(\log 2)\frac{\pi }{4}\\ \Rightarrow \frac{1}{a}{\int }_0^{\pi /4} \log (1+\tan u)du=\frac{1}{a}(\log 2)\frac{\pi }{8}\end{array}$