With the substitution $\arctan x=t$ the integral takes the form$$ I=\int_{0}^{\frac{\pi}{4}} \ln (1+\tan t) d t . $$This we already computed in the previous problem. ("Happiness is longing for repetition," says M. Kundera.) So the answer to the problem is $\frac{\pi}{8} \ln 2$.
Find$$ \int_{0}^{1} \frac{\ln (1+x)}{1+x^{2}} d x .$$