Developing $\int_0^1 \frac{1+x}{1-x^3} \ln\left(\frac{1}{x}\right)dx$ into series I'm having trouble developing this integral into a series:
$$\int_0^1 \frac{1+x}{1-x^3} \ln\left(\frac{1}{x}\right)dx$$
Anyone has an idea?

Make a substitution $x=e^{-y}$:
$$\begin{align}\int_0^1 dx\: \frac{1+x}{1-x^3} \ln\left(\frac{1}{x}\right) &= \int_0^{\infty} dy \: y \,e^{-y} \frac{1+e^{-y}}{1-e^{-3 y}}\\ &= \int_0^{\infty} dy \: y  (e^{-y}+e^{-2 y}) \sum_{k=0}^{\infty} e^{-3 k y}\\&=\sum_{k=0}^{\infty} \int_0^{\infty} dy \: y  (e^{-(3 k+1)y}+e^{-(3 k+2) y})\\&= \sum_{k=0}^{\infty} \left [ \frac{1}{(3 k+1)^2}+\frac{1}{(3 k+2)^2}\right]\\ &=\sum_{k=0}^{\infty} \left [ \frac{1}{( k+1)^2}-\frac{1}{(3 k+3)^2}\right]\\&=\frac{\pi^2}{6} - \frac{1}{9}\frac{\pi^2}{6}\\&=\frac{4 \pi^2}{27}\end{align}$$
In general,
$$\int_0^1 dx \frac{1-x^{n-1}}{(1-x)(1-x^n)} \left[\ln{\left(\frac{1}{x}\right)}\right]^{m-1} =  \left(1-\frac{1}{n^m}\right) \Gamma(m) \zeta(m)$$
