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This question already has an answer here:

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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marked as duplicate by lab bhattacharjee calculus Sep 4 '15 at 7:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let me try.

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, $$I = \frac{\pi}{8}\ln 2.$$

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