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
 A: Just to simplify the things, make the change of variables $s=2x-1$. The integral then reduces to
$$I=2\int_{-1}^1\frac{\ln\frac{3+s^2}{4}}{1-s^2}ds.\tag{1}$$
The antiderivative of any expression of the type $\displaystyle\frac{\ln P(x)}{Q(x)}$ is computable in terms of dilogarithms, essentially due to 
$$\displaystyle \int\frac{\ln(a-x)}{x+b}dx=\mathrm{Li}_2\left(\frac{a-x}{a+b}\right)+\ln(a-x)\ln\frac{x+b}{a+b}.\tag{2}$$ 
Hence the answer can be certainly expressed in terms of dilogarithm values.
Let us spell this out more explicitly. It is convenient to integrate once by parts and rewrite (1) as
\begin{align}I=&-\int_{-1}^1\frac{2s}{3+s^2}\ln\frac{1+s}{1-s}ds=
4\Re\int_{-1}^1\frac{\ln(1-s)}{s+i\sqrt3}ds.
\end{align}
Applying (2), this reduces to
$$I=-4\Re\,\mathrm{Li}_2\left(e^{i\pi/3}\right)=-\frac{\pi^2}{9},$$
where at the last step we have used that for $z\in(0,1)$ one has
$$\Re\,\mathrm{Li}_2\left(e^{2i\pi z}\right)=\pi^2\left(z^2-z+\frac16\right).$$
A: 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}
A: 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 ($*$).
