Definite Integral $\int_0^1\frac{\ln(x^2-x+1)}{x^2-x}\,\mathrm{d}x$ $$\int_0^1\frac{\ln(x^2-x+1)}{x^2-x}\,\mathrm{d}x$$
WA gives $\pi^2/9$

Mhenni has struck first with the approach I have taken, but I would like to elaborate.  Again, the integrand may be Taylor expanded:
$$\begin{align}-\int_0^1 dx \frac{\log{[1-(x-x^2)]}}{x-x^2}  &= \sum_{n=0}^{\infty} \frac1{n+1} \int_0^1 dx \, x^n (1-x)^n\\ &= \sum_{n=0}^{\infty} \frac1{n+1} \frac{n!^2}{(2 n+1)!}\\ &=2 \sum_{n=0}^{\infty} \frac1{(2 n+2) (2 n+1) \binom{2 n}{n}} \end{align}$$
It turns out that 
$$\frac{\arcsin{x}}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{2^{2 n} x^{2 n+1}}{(2 n+1) \binom{2 n}{n}} $$
So the sum in question is simply
$$4 \int_0^1 dx \frac{\arcsin{(x/2)}}{\sqrt{1-x^2/4}} = 8 \int_0^{\pi/6} d\theta \, \theta = \frac{\pi^2}{9}$$
as was to be shown.
