Prove that $\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\, dx=\frac{\pi}{8}\log 2$

Asked Sep 13 '15 at 14:53

Active Sep 13 '15 at 14:53

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Prove that $\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\, dx=\frac{\pi}{8}\log 2$
I have tried to sole the problem, but failed. Any one can help me to sole the problems.

Here, I have tried to solve this using integration by parts :
Let $I=\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\, dx$. Then
$I=\left[ \tan^{-1}x\,\log(1+x)\right]_0^1-\int_{0}^{1} \frac{\tan^{-1}x}{1+x}\, dx$

How can I integrate the second term?

asked Sep 13 '15 at 14:53

MKSar

The function you are integrating has an indefinite integral which has no closed form, so the method you are trying to apply will probably not work. Try using some of the tricks for calculating improper integrals... – 5xum Sep 13 '15 at 14:57

Prove that $\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\, dx=\frac{\pi}{8}\log 2$

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