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.

You can use, since you're in the range $[0, 1]$, the logarithm expansion
$$\ln(x^2 - x + 1) = \sum_{k = 1}^{+\infty} \frac{(-1)^k}{k} (x^2-x)^k$$
together with the fact that $x - x^2 = -(x^2 - x)$, hence the integration becomes
$$\int_1^0$$
due to the minus sign, incorporated into the extrema.
After that, treat the fraction
$$\frac{(x^2-x)^{k}}{x^2 - x} = (x^2-x)^{k-1}$$
With the binomial theorem, which transforms it into
$$\sum_{j = 0}^{k-1} \binom{k-1}{j} (-1)^{k-1-j}$$
Giving you the integral 
$$\sum_{k = 1}^{+\infty} \frac{(-1)^k}{k} \sum_{j = 0}^{k-1} \binom{k-1}{j} (-1)^{k-1+j} \int_1^0 x^{k-1+j} dx$$
Which is trivial.
The final result in terms of series:
$$-\sum_{k = 1}^{+\infty} \frac{(-1)^{2k-j}}{k} \sum_{j = 0}^{k-1} \binom{k-1}{j} \left(\frac{1}{k+j}\right)$$
The series can be summed and the final result is numerical 
$$-\frac{\pi^2}{9}$$
