<s>[INST] 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
 [/INST] 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}
</s>