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 $\pi^2/9$
 A: 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.
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$\ds{\int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x^{2} - x}\,\dd x:\ {\large ?}}$

Roots of $\ds{x^{2} - x + 1 = 0}$ are given by
  $\ds{\quad x_{\rm r} \equiv {1 + \root{3}\ic \over 2}\quad}$ and
  $\ds{\quad{1 \over x_{\rm r}} = x_{\rm r}^{*}}$.

\begin{align}&\color{#c00000}{%
\int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x^{2} - x}\,\dd x}
=-\int_{0}^{1}\ln\pars{x^{2} - x + 1}\pars{{1 \over x} + {1 \over 1 - x}}\,\dd x
\\[3mm]&=-2\int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x}\,\dd x
=-2\bracks{\int_{0}^{1}{\ln\pars{1 - x/x_{\rm r}} \over x}\,\dd x+
\int_{0}^{1}{\ln\pars{1 - x/x_{\rm r}^{*}} \over x}\,\dd x}
\\[3mm]&=-2\bracks{\int_{0}^{1/x_{\rm r}}{\ln\pars{1 - x} \over x}\,\dd x+
\int_{0}^{1/x_{\rm r}*}{\ln\pars{1 - x} \over x}\,\dd x}
=2\bracks{{\rm Li}_{2}\pars{1 \over x_{\rm r}} +
{\rm Li}_{2}\pars{x_{\rm r}}}
\end{align}
where $\ds{{\rm Li}_{2}\pars{z}}$ is the Dilogarithm Function.
Note that $\ds{{\rm Li}_{1}\pars{z} = -\ln\pars{1 - z}}$ and
$\ds{{\rm Li}_{2}\pars{z} = \int_{0}^{z}{{\rm Li}_{1}\pars{t} \over t}\,\dd t}$.

By using the Dilogarithm Inversion Formula
  $\ds{{\rm Li}_{2}\pars{z} + {\rm Li}_{2}\pars{1 \over z}
     =-\,{\pi^{2} \over 6} - \half\,\ln^{2}\pars{-z}}$ where $\ds{z \not\in \left[\vphantom{\Large A}0, 1\right)}$:
  \begin{align}&\color{#c00000}{%
\int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x^{2} - x}\,\dd x}
=2\bracks{-\,{\pi^{2} \over 6} - \half\,\ln^{2}\pars{-1 - \root{3}\ic \over 2}}
\\[3mm]&=2\bracks{-\,{\pi^{2} \over 6} - \half\,\pars{-\,{2\pi \over 3}}^{2}}
\end{align}

$$
\color{#66f}{\large%
\int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x^{2} - x}\,\dd x = {\pi^{2} \over 9}}
$$
A: Here is an approach which is based on the Taylor series and the beta function
$$ I = \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} \int_{0}^{1}\frac{(x^2-x)^k}{(x^2-x)}dx = \sum_{k=0}^{\infty}\frac{1}{k} \int_{0}^{1}{x^{k-1}(1-x)^{k-1}}dx $$
$$ = \sum_{k=0}^{\infty}\frac{1}{k} \beta(k,k) = \sum_{k=0}^{\infty}\frac{1}{k} \frac{\Gamma(k)\Gamma(k)}{\Gamma(2k)} = 4(\sin^{-1}(1/2))^2 \sim 1.096622711. $$
A: Let me present you the simplest approach I could even think about. At start, I have to say I am using a method suggested earlier in comments by @Lucian . So it begins by computation of this kind of integrals, by expanding the integral in power series : 
$$I_n = \int_0^1 \frac{\ln(1-x^n)}{x} \;\mathrm{d}x= -\int_0^1 \sum_{k=1}^{\infty} \frac{x^{nk-1}}{k} \;\mathrm{d}x = -\sum_{k=1}^{\infty} \frac{1}{nk^2}=-\frac{\zeta{(2)}}{n} $$
Next, using the fact that,
$$\frac{1}{x^2-x}=-\frac{1}{1-x}-\frac{1}{x}$$
And $x^2-x+1$ is symmetric inder transformation $x \to 1-x$ we get for our original integral :
$$I=-2\int_0^1\frac{\ln\left(x^2-x+1\right)}{x}\;\mathrm{d}x=-2\int_0^1\frac{\ln\left(\frac{1-x^6}{1-x^3}\cdot\frac{1-x}{1-x^2}\right)}{x}\;\mathrm{d}x = -2\left(I_6-I_3-I_2+I_1\right)=2\zeta(2)\left(\frac16-\frac13-\frac12+1\right)=\frac23\zeta(2)=\frac{\pi^2}{9}$$ 
