Definite Integral $\int_{0}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx$ for Logarithm and Algebraic. 
Evaluate
  $$\int_{0}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx$$

I have been thinking for this question quite some time, I've tried all the methods I have learnt but still getting nowhere. Hope that someone can explain it for me. Thanks in advance.

Note that $t=x(1-x)$ is increasing in $(0,1/2)$ and decreasing in $(1/2,1)$. So
$$ x_\pm(t)=\frac12(1\pm\sqrt{1-4t}) $$
from which $$x_-(t):[0,\frac14]\to[0,1\frac12], x_+(t):[0,\frac14]\to[0,\frac12].$$
Thus
\begin{eqnarray}
&&\int_{0}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx\\
&=&\int_{0}^{1/2}\frac{\ln(x^2-x+1)}{x-x^2}dx+\int_{1/2}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx\\
&=&\int_{0}^{1/2}\frac{\ln(x^2-x+1)}{x-x^2}dx+\int_{1/2}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx\\
&=&2\int_{0}^{1/4}\frac{\ln(1-t)}{t\sqrt{1-4t}}dt\\
&=&-2\int_{0}^{1/4}\sum_{n=1}^\infty\frac1n \frac{t^{n-1}}{\sqrt{1-4t}}dt\\
&=&-2\sum_{n=1}^\infty\frac{\sqrt{\pi}4^{-n}(n-1)!}{n\Gamma(n+\frac12)}\\
&=&-4\arcsin^2(\frac12)\\
&=&-\frac{\pi^2}{9}.
\end{eqnarray}
Here
$$ \int_0^{1/4}\frac{t^{n-1}}{\sqrt{1-4t}}dt=\frac{\sqrt\pi4^{-n}(n-1)!}{\Gamma(n+\frac12)},\sum_{n=1}^\infty\frac{\sqrt{\pi}(n-1)!}{n\Gamma(n+\frac12)}x^{2n}=2\arcsin^2(x) $$
are used.
