$\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}\frac{\ln \left(x+1\right)}{x^2+1}\mathrm{dx}=?$

we substitute

$x=\tan t=\frac{\sin t}{\cos t}$

$\mathrm{dx}=\frac{\cos t·\; <!--\; middle\; dot\; -->\cos t+\sin t·\; <!--\; middle\; dot\; -->\sin t}{{\cos }^{2}t}=\frac{1}{{\cos }^{2}t}={\sec }^{2}t\mathrm{dt}$

Also
note that

${\tan }^{2}t+1=\frac{{\sin }^{2}t}{{\cos }^{2}t}+1={\sec }^{2}t$

After substituting new variable and change integral bounds accordingly we have

$\underset{0}{\overset{\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}}{\int \; <!--\; integral\; -->}}\frac{\ln \left(\tan t+1\right)}{{\tan }^{2}t+1}·\; <!--\; middle\; dot\; -->{\sec }^{2}t\mathrm{dt}=\underset{0}{\overset{\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}}{\int \; <!--\; integral\; -->}}\ln \left(\tan t+1\right)$

So now, how can we solve this integral?

If we have an integral in general we have following identity

$\underset{0}{\overset{a}{\int \; <!--\; integral\; -->}}f\left(x\right)\mathrm{dx}=\underset{0}{\overset{a}{\int \; <!--\; integral\; -->}}f\left(a-x\right)\mathrm{dx}$

which can be shown by changing variable x= a-u and rename the variable to × again

$\underset{0}{\overset{\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}}{\int \; <!--\; integral\; -->}}\ln \left(\tan t+1\right)=\underset{0}{\overset{\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}}{\int \; <!--\; integral\; -->}}\ln \left(\tan \left(\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}-t\right)+1\right)$

Now, we have

$\tan \left(\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}-t\right)=\frac{\tan \frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}-\tan t}{1+\tan \frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}·\; <!--\; middle\; dot\; -->\tan t}=\frac{1-\tan t}{1+\tan t}$

we can
do the
summation

$\underset{0}{\overset{\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}}{\int \; <!--\; integral\; -->}}\ln \left(\frac{1-\tan t}{1+\tan t}+1\right)\mathrm{dt}=\underset{0}{\overset{\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}}{\int \; <!--\; integral\; -->}}\ln \left(\frac{2}{1+\tan t}\right)\mathrm{dt}=\underset{0}{\overset{\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}}{\int \; <!--\; integral\; -->}}\ln \left(2\right)\mathrm{dt}-\underset{0}{\overset{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}/4}{\int \; <!--\; integral\; -->}}\ln \left(\tan t+1\right)\mathrm{dt}$

$\underset{0}{\overset{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}/4}{\int \; <!--\; integral\; -->}}\ln \left(\tan t+1\right)\mathrm{dt}=\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}\frac{\ln \left(x+1\right)}{x^2+1}\mathrm{dx}$

Remember this was
the integral we started
with so finally

$\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}\frac{\ln \left(x+1\right)}{x^2+1}\mathrm{dx}=\frac{\ln \left(2\right)·\; <!--\; middle\; dot\; -->\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}}{2}=\frac{\ln 2.\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{8}$