How to evaluate $\int_{0}^{1} \frac{\ln x}{x+1} dx$ I want to evaluate:
$$\int_{0}^{1} \frac{\ln x}{x+1} dx$$
If I was asked I would to evaluate:
$$\int_{0}^{1} \frac{\ln x}{x-1} dx$$
That would be easy because if I use the Taylor series for $\ln x$ centered at $1$ then things will cancel out and leave me with a easy integral.
So how about this integral, I'm guessed to use  the Taylor series centered around $-1$. But even with that thought in mind,  it does not take me anywhere because $\ln (-1)$ is not defined. Can someone help.

We know that $$\int_0^1 x^a~dx=\frac{1}{a+1}$$
Then we can differentiate with respect to $a$:
$$\int_0^1 x^a \ln x ~dx=-\frac{1}{(a+1)^2}$$
Now we can use the geometric series:
$$\sum_{a=0}^\infty (-1)^a x^a=\frac{1}{1+x},~~~|x|<1$$
$$\sum_{a=0}^\infty (-1)^a  \int_0^1 x^a \ln x ~dx=- \sum_{a=0}^\infty (-1)^a  \frac{1}{(a+1)^2}$$
(Now we interchange integration and summation on the left hand side)
$$\int_0^1 \frac{\ln x}{1+x} ~dx=-\sum_{a=0}^\infty \frac{(-1)^a}{(a+1)^2}=\sum_{a=1}^\infty \frac{(-1)^a}{a^2}=-\frac{\pi^2}{12}$$
