closed form for $\int_0^1\frac{\mathrm{Li}_s(x-x^2)}{x-x^2}\mathrm dx$ I am trying to evaluate 
$$F(s)=\sum_{n\geq1}\frac1{n^{s+1}{2n\choose n}}$$
I started off by noting that $$\frac1{n{2n\choose n}}=\frac12\int_0^1\left[x-x^2\right]^{n-1}\mathrm dx$$
So
$$F(s)=\int_0^1\sum_{n\geq1}\frac{\left[x-x^2\right]^{n-1}}{n^s}\mathrm dx$$
And when we recall the definition of the polylogarithm function
$$\mathrm{Li}_s(z)=\sum_{n\geq1}\frac{z^n}{n^s}$$
It becomes apparent that 
$$F(s)=\int_0^1\frac{\mathrm{Li}_{s}(x-x^2)}{x-x^2}\mathrm dx$$
Which I do not know how to deal with. Could I have some help evaluating this integral? Thanks.
 A: An expression as a generalized hypergeometric function can be directly obtained by expressing the ratio of two consecutive terms in the series which defines $F(s)$ as a rational fraction in $n$ allowing thus to express the series as a generalized hypergeometric series. 
Alternatively, starting from the integral, as remarked by  @mrtaurho
\begin{align}
   F(s)&=\int_0^{1/2}\frac{\mathrm{Li}_{s}(x-x^2)}{x-x^2}\,dx+\int_{1/2}^1\frac{\mathrm{Li}_{s}(x-x^2)}{x-x^2}\,dx\\
   &=2\int_0^{1/2}\frac{\mathrm{Li}_{s}(x-x^2)}{x-x^2}\,dx\\
   &=2\int_0^{1/4}\frac{\mathrm{Li}_{s}(t)}{t}\frac{dt}{\sqrt{1-4t}}\\
   &=2\int_0^{1}\frac{\mathrm{Li}_{s}(\frac{u}{4})}{u}\frac{du}{\sqrt{1-u}}
  \end{align}
(we changed $x\to 1-x$ n the second integral of the first expression, then $t=x(1-x)$ and finally $t=u/4$). Now, if $s$ is an integer, we use the  representation in terms of the hypergeometric function (here)
\begin{equation}
 \mathrm{Li}_s(z)=z\,_{s+1}F_s\left( 1,1,\ldots,1;2,2,\ldots,2;z \right)
\end{equation} 
to obtain
\begin{equation}
 F(s)=\frac{1}{2}\int_0^{1}\left( 1-u \right)^{-1/2}{}_{s+1}F_s\left( 1,1,\ldots,1;2,2,\ldots,2;\frac{u}{4} \right)\,du
\end{equation} 
Then, from the tabulated integral (DLMF),
\begin{equation}
 {{}_{p+1}F_{q+1}}\left({a_{0},\dots,a_{p}\atop b_{0},\dots,b_{q}};z\right)=%
\frac{\Gamma\left(b_{0}\right)}{\Gamma\left(a_{0}\right)\Gamma\left(b_{0}-a_{0%
}\right)}\int_{0}^{1}t^{a_{0}-1}(1-t)^{b_{0}-a_{0}-1}{{}_{p}F_{q}}\left({a_{1}%
,\dots,a_{p}\atop b_{1},\dots,b_{q}};zt\right)\mathrm{d}t
\end{equation} 
valid for $\Re b_0>\Re a_0>0$ if $p\le q$ and, additionally, $\left|\mathrm{ph}(1-z)\right|<\pi$ if $p=q+1$. Here $a_0=1,b_0=3/2,z=1/4$, thus
\begin{equation}
 F(s)={}_{s+2}F_{s+1}\left( 1,1,\ldots,1;\frac{3}{2},2,2,\ldots,2;\frac{1}{4} \right)
\end{equation} 
