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

Differentiation under the integral sign can be applied to this.
Consider 
\begin{align}
I(a)&=\int_0^1 \, \frac{\ln{(1-a\,(x-x^2))}}{x-x^2}\, dx \tag 1\\
\frac{\partial}{\partial a} I(a)&=\int_0^1\, -\frac{1}{1-a\,(x-x^2)}\, dx \\
&= -\frac{4 \, \sqrt{-y^{2} + 4 \, y} \arctan\left(\frac{\sqrt{-y^{2} + 4 \, y}}{y - 4}\right)}{y^{2} - 4 \, y}\tag 2
\end{align}
Integrating $(2)$ gives us (I used Sage for that)
\begin{align}
I(a)=\sqrt{a} \sqrt{-a + 4} \arcsin\left(\frac{1}{2} \, a - 1\right) + 2 \, \arcsin\left(\frac{1}{2} \, a - 1\right)^{2} - 4 \, \arctan\left(\frac{{\left(a - 2\right)} \sqrt{a} \sqrt{-a + 4}}{a^{2} - 4 \, a}\right) \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right) + 4 \, \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right)^{2} + 4 \, \arcsin\left(\frac{1}{2} \, a - 1\right) \arctan\left(\frac{\sqrt{-a^{2} + 4 \, a}}{a - 4}\right) - \sqrt{-a^{2} + 4 \, a} \arcsin\left(\frac{1}{2} \, a - 1\right) + C \\
\tag 3
\end{align}
Setting $a=0$ in $(2)$ and $(3)$, $C=-\frac{\pi^2}{2}$, hence:
\begin{align}
I(a)=\sqrt{a} \sqrt{-a + 4} \arcsin\left(\frac{1}{2} \, a - 1\right) + 2 \, \arcsin\left(\frac{1}{2} \, a - 1\right)^{2} - 4 \, \arctan\left(\frac{{\left(a - 2\right)} \sqrt{a} \sqrt{-a + 4}}{a^{2} - 4 \, a}\right) \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right) + 4 \, \arctan\left(\frac{\sqrt{-a + 4}}{\sqrt{a}}\right)^{2} + 4 \, \arcsin\left(\frac{1}{2} \, a - 1\right) \arctan\left(\frac{\sqrt{-a^{2} + 4 \, a}}{a - 4}\right) - \sqrt{-a^{2} + 4 \, a} \arcsin\left(\frac{1}{2} \, a - 1\right)-\frac{\pi^2}{2}
\end{align}
Therefore, the required answer is
$$
I(1)=-\frac{\pi^2}{9}\approx -1.09662271123215
$$
=== Update ===
There is a simpler general form as I suspected:
Instead of $(1)$, consider $I(a)$ to be
\begin{align}
I(a)&=\int_0^1 \, \frac{\ln{(1+a\,(x-x^2))}}{x-x^2}\, dx
\end{align}
Applying the differentiation under the integral sign now yields a nice simpler form:
\begin{align}
I(a)&=\ln\left( \frac{a+2 -\sqrt{a^{2} + 4 \, a}}{2}\right)^{2}
\end{align}
