Integrate $\int_{0}^{1}\frac{\ln(x)}{x+1} dx $

Question

How can I integrate this? $\int_{0}^{1}\frac{\ln(x)}{x+1} dx $

I've seen this but I failed to apply it on my problem.

Could you give some hint?

EDIT : From hint of @H.H.Rugh, I've got $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2}$, since $\int_{0}^{1}x^{n}\ln(x)dx = (-1)\frac{1}{(n+1)^2}$. How can I proceed this calculation hereafter?

Answer 1

Hint: $\frac{1}{1+x}=1-x+x^2-x^3...$ Then integrate term by term using partial integration.

Answer 2

You can derivate $Li_2(-x)+(\ln x)(\ln (x+1))+C$ where $\displaystyle \frac{d}{dz}Li_2(z)=-\frac{\ln(1-z)}{z}$ and you will see how to integrate.

Hint: $\displaystyle\enspace Li_s(z):=\sum\limits_{k=1}^\infty \frac{z^k}{k^s}$ and $\displaystyle\enspace Li_2(1)=\frac{\pi^2}{6}$ .