So we got this problem $$8\int_{0}^{1}\frac{\log (1+x)}{1+x^2}dx$$ I have been stuck on this problem for days basically I tried everything I could think of to solve this integral i tried substituting $$x=\tan p$$ and integral became $$8\int_{0}^{1}\frac{\log (1+\tan p)}{1+\tan^2 p}(\sec^2 p)\, dp$$ then I was stuck at $$8\int_{0}^{1}{\log (1+\tan p)}\,dp $$ now nowhere to go from here can you suggest me another way to approach this problem ?
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http://math.stackexchange.com/questions/220746/integrating-frac-log1x1x2/220754#220754 – lab bhattacharjee Jan 27 '14 at 12:16
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I collected a few integrals of this type http://math.stackexchange.com/questions/542477/further-our-knowledge-of-a-certain-class-of-integral-involving-logarithms – Bennett Gardiner Jan 27 '14 at 12:32
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Hint: Use $$\log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}$$ and argue why one can interchange integral and sum.
J.R.
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