$$\int_0^1 dx\: \frac{1+x}{1-x^3} \ln\left(\frac{1}{x}\right)$$

This one is relatively simple and wouldn’t normally make it to this blog, but the result is easily generalizable to a very cool, highly nontrivial result.

To start, 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,

$$\int_0^1 dx \frac{1-x^{n-1}}{(1-x)(1-x^n)} \left[\ln{\left(\frac{1}{x}\right)}\right]^{m-1} = \left(1-\frac{1}{n^m}\right) \Gamma(m) \zeta(m)$$

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