# Integral of $8\int_{0}^{1}\frac{\log (1+x)}{1+x^2}dx$ [duplicate]

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 ?
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.