# Solve $I=\int_0^1\frac{ln(1+x)}{1+x^2}dx$ [duplicate]

Solve $$I=\int_0^1\frac{ln(1+x)}{1+x^2}dx.$$ After let $x=\tan t$, $I=\int_0^{\pi/4}ln(1+\tan t)dt$ and I stuck here.

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Let $v = \frac{\pi}{4}-t$. We have
$$I = \int_0^{\frac{\pi}{4}} \ln (1 + \tan(\frac{\pi}{4}-v))dv = \int_0^{\frac{\pi}{4}} \ln \left(1 + \frac{1-\tan v}{1 + \tan v}\right)dv = \int_0^{\frac{\pi}{4}} \ln \left(\frac{2}{1+\tan v}\right)dv \\ = \int_0^{\frac{\pi}{4}} \ln 2 - \int_0^{\frac{\pi}{4}} \ln \left(1+\tan v\right)dv = \frac{\pi}{4}\ln 2 - I.$$
So, $$I = \frac{\pi}{8}\ln 2.$$