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.

Convert the integral to the easier one via
\begin{align}
\int_{0}^{1} \frac{\ln x}{x+1} dx &= 
 \int_{0}^{1} \frac{\ln x}{1-x} dx - \int_{0}^{1} \frac{2x\ln x}{1-x^2} dx, \>\>\>\>\>t=x^2\\
&= \int_{0}^{1} \frac{\ln x}{1-x} dx -\frac12 \int_{0}^{1} \frac{\ln t}{1-t} dt \\
&= \frac12 \int_{0}^{1} \frac{\ln t}{1-t} dt=\frac12\cdot \frac{\pi^2}{6}=\frac{\pi^2}{12}
\end{align}
