<s>[INST] 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.
 [/INST] 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
$$\int^1_0 \frac{\ln \left (x^2 - x + 1 \right )}{x - x^2} \, dx = -\frac{\pi^2}{9}.$$
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