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.

If you want to avoid series altogether one could always try Feynman's trick of differentiating under the integral sign.
Consider
$$I(a) = \int^1_0 \frac{\ln \left [a(x^2 - x) + 1 \right ]}{x - x^2} \, dx, \quad a \geqslant 0.$$
Note that $I(0) = 0$ and we require $I(1)$. 
Differentiating under the integral sign with respect to the parameter $a$ we have
$$I'(a) = - \int^1_0 \frac{dx}{a(x^2 - x) + 1}.$$
Completing the square in the denominator before integrating we have
\begin{align*}
I'(a) &= -\frac{1}{a} \int^1_0 \frac{dx}{\left (x - \frac{1}{2} \right )^2 + \left (\frac{1}{2} \sqrt{\frac{4 - a}{a}} \right )^2}\\
&= -\frac{2}{a} \sqrt{\frac{a}{4 - a}} \left \{\tan^{-1} \left [2 \sqrt{\frac{a}{4 - a}} \left (x - \frac{1}{2} \right ) \right ]\right \}^1_0\\
&= -\frac{4}{a} \sqrt{\frac{a}{4 - a}} \tan^{-1} \sqrt{\frac{a}{4 - a}}.
\end{align*}
Thus
$$I(1) = -4 \int^1_0 \sqrt{\frac{a}{4 - a}} \tan^{-1} \sqrt{\frac{a}{4 - a}} \cdot \frac{da}{a}.$$
Letting
$$u^2 = \frac{a}{4 - a},$$
observing that
$$a = \frac{4u^2}{1 + u^2},$$
we have
$$da = \frac{8 u}{(1 + u^2)^2} \, du,$$
while for the limits of integration: $(0,1) \mapsto (0,1/\sqrt{3})$. Thus
$$I(1) = - 8 \int^{1/\sqrt{3}}_0 \frac{\tan^{-1} u}{1 + u^2} \, du.$$
This last integral can be readily found by setting $t = \tan^{-1} u$. Doing so yields
$$I(1) = -8 \int^{\pi/6}_0 t \, dt = -8 \left [\frac{t^2}{2} \right ]^{\pi/6}_0,$$
or
$$\int^1_0 \frac{\ln \left (x^2 - x + 1 \right )}{x - x^2} \, dx = -\frac{\pi^2}{9}.$$
