Calculate $\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$ I am trying to calculate:
$$\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$$
I am not looking for an answer but simply a nudge in the right direction. A strategy, just something that would get me started.
So, after doing the Taylor Expansion on the $\ln(1-x+x^2)$ ig to the following: Let $x=x-x^2$ then $\ln(1-x)$ then,
\begin{align*}
=&-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-...\\
=&-(x-x^2)-\frac{(x-x^2)^2}{2}-...\\
=&-x(1-x)+\frac{x^2}{2}(1-x)^2-\frac{x^3}{3}(1-x)^3\\
\text{thus the pattern is:}\\
=&\frac{x^n(1-x)^n}{n}
\end{align*} 
Am I right?
Then our Integral would be: $$\sum_{n=0}^{\infty} \frac{1}{n+1} \int_0^1 x^n(1-x)^n$$
Am I on the right track? Suggestions, tips, comments?
$\underline{NEW EDIT:}$
SO after integrating the function I got the following after a couple of iterations:
\begin{align*}
\frac{n(n-1)...1}{(n+1)(n+2)...(2n)}\int_0^1 x^{2n} dx
\end{align*}
This shows a pattern:
\begin{align*}
=&\frac{(n!)^2}{(2n)!} (\frac{1}{2n+1})\\
=& \frac{(n!)^2}{(2n+1)!}
\end{align*}
So my question is, what to do from here. I have done all this but still have no clue how to actually solve the integral. Can somebody shed some light on this!
Thanks

Here is a relatively simple way that only relies on knowing the following Maclaurin series expansion for the square of the inverse sine function (for several different proofs of this, see here)
$$(\sin^{-1} x)^2 = \frac{1}{2} \sum_{n = 1}^\infty \frac{(2x)^{2n}}{n^2 \binom{2n}{n}}, \qquad |x| \leqslant 1.$$
Note that if we set $x = 1/2$ one obtains:
$$\sum_{n = 1}^\infty \frac{1}{n \binom{2n}{n}} = \frac{\pi^2}{18}. \qquad (*)$$
Now
\begin{align}
\int_0^1 \frac{\ln (1 - x + x^2)}{x - x^2} \, dx &= \int_0^1 \frac{\ln [1 - (x - x^2)]}{x - x^2} \, dx\\
&= -\int_0^1 \sum_{n = 1}^\infty \frac{(x - x^2)^n}{n} \frac{dx}{x - x^2} \tag1\\
&= -\sum_{n = 1}^\infty \frac{1}{n} \int_0^1 (x - x^2)^{n - 1} \, dx \tag2\\
&= -\sum_{n = 1}^\infty \frac{1}{n} \int_0^1 x^{n - 1} (1 - x)^{n - 1} \, dx\\
&= -\sum_{n = 1}^\infty \frac{1}{n} \operatorname{B} (n,n) \tag3\\
&= -\sum_{n = 1}^\infty \frac{1}{n} \frac{\Gamma (n) \Gamma (n)}{\Gamma (2n)} \tag4\\
&= -\sum_{n = 1}^\infty \frac{1}{n} \frac{(n - 1)! (n - 1)!}{(2n - 1)!}\\
&= -2\sum_{n = 1}^\infty \frac{1}{n} \frac{(n!)^2}{(2n)!}\\
&= -2 \sum_{n = 1}^\infty \frac{1}{n \binom{2n}{n}}\\
&= -\frac{\pi^2}{9} \tag5
\end{align}
Explanation
(1): Maclaurin series expansion for $\ln (1 - z)$.
(2): The dominated convergence theorem allows the summation and integration signs to be interchanged. 
(3): Integral representation for the beta function.
(4): Using the property $\operatorname{B} (x,y) = \frac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$.
(5): Using the result given above in ($*$).
