Find the value of $A+B+C$ in the following question? 
Find the value of $A+B+C$ such that
  $$S=\sum_{n=1}^{n=\infty}\dfrac{1}{n^2\binom{2n}{n}}=\dfrac{A}{B}\zeta(C).$$

I solved it as follows: by using formula 
 $$\sum_{n=1}^{n=\infty}\dfrac{1}{n\binom{2n}{n}}=\beta(n+1,n)=\beta(n,n+1)$$
$$S=\sum_{n=1}^{n=\infty}\dfrac{1}{n}\beta(n+1,n)$$
then i used definition of beta function to rewrite above sum as follows
  $$S=\sum_{n=1}^{n=\infty}\dfrac{1}{n}\int_{0}^{1} x^n(1-x)^{n-1}dx$$
then after changing order of integration and summation i again modified above as
$$S=\displaystyle\int_{0}^{1}\sum_{n=1}^{n=\infty}\left[\dfrac{x^n(1-x)^{n-1}}{n}\right]$$
 from here i don't know how to proceed and relate it to zeta function (i don't no much about zeta function). Any help would be appreciated.
 A: More generally, the following power series expansion holds (see for example HERE): for $|x|\leq 1$
$$2(\arcsin(x))^2=\sum_{n=1}^{\infty} \frac{(2x)^{2n}}{n^2\binom{2n}{n}}.$$
Hence, for $x=1/2$, we find that
$$\sum_{n=1}^{\infty} \frac{1}{n^2\binom{2n}{n}}=2(\arcsin(1/2))^2=2\left(\frac{\pi}{6}\right)^2=\frac{1}{3}\zeta(2).$$
Finally it easy to obtain $A+B+C=1+3+2=6$.
P.S. See also How to prove by arithmetical means that $\sum\limits_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} =\frac{1}{3}\sum\limits_{k=1}^{\infty}\frac{1}{k^{2}}$
A: The famous identity $$\zeta(2)=\sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}}$$
can be proved in many ways: creative telescoping, complex analysis, Lagrage's inversion theorem or Legendre polynomials, just to mention a few of them. Have a look at the first section of my notes.
A self-contained proof:
$$\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{n^2\binom{2n}{n}}=\sum_{n\geq 1}\frac{(n-1)!^2}{(2n)!}&=&\sum_{n\geq 1}\frac{\Gamma(n)^2}{2n\,\Gamma(2n)}\\&=&\sum_{n\geq 1}\frac{B(n,n)}{2n}\\&=&\sum_{n\geq 1}\frac{1}{2n}\int_{0}^{1}x^{n-1}(1-x)^{n-1}\,dx\\&=&-\frac{1}{2}\int_{0}^{1}\frac{\log(1-x+x^2)}{x(1-x)}\,dx\end{eqnarray*}$$
and now it is enough to notice that $\frac{1}{x(1-x)}=\frac{1}{x}+\frac{1}{1-x}$ and that $1-x+x^2$ is a cyclotomic polynomial, in order to exploit this lemma:
$$ \int_{0}^{1}\frac{\log\Phi_n(x)}{x}\,dx = \frac{\zeta(2)(-1)^{\omega(n)+1}\varphi(n)\,\text{rad}(n)}{n^2}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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I'll start from de
  Jack D'Aurizio "integral expression":

\begin{align}
S & = -\,{1 \over 2}\int_{0}^{1}{\ln\pars{1 - x + x^{2}} \over
x\pars{1 - x}}\,\dd x =
-\,{1 \over 2}\int_{0}^{1}
{\ln\pars{1 - x + x^{2}} \over x}\,\dd x - {1 \over 2}\
\overbrace{\int_{0}^{1}{\ln\pars{1 - x + x^{2}} \over 1 - x}\,\dd x}
^{\ds{\mbox{lets}\ x\ \mapsto\ 1 - x}}
\\[5mm] & =
-\int_{0}^{1}{\ln\pars{1 - x + x^{2}} \over x}\,\dd x
\,\,\,\stackrel{\mrm{IBP}}{=}\,\,\,
\int_{0}^{1}\ln\pars{x}\,{2x - 1 \over x^{2} - x + 1}\,\dd x =
\int_{0}^{1}\ln\pars{x}\,{2x - 1 \over \pars{x - r}\pars{x - \bar{r}}}\,\dd x
\end{align}

where
  $\ds{r \equiv {1 + \root{3}\ic \over 2} = \exp\pars{{\pi \over 3}\,\ic}}$.

Then,
\begin{align}
S & =
\int_{0}^{1}\ln\pars{x}\bracks{{2r - 1 \over \pars{r - \bar{r}}\pars{x - r}} -{2\bar{r} - 1 \over \pars{r - \bar{r}}\pars{x - \bar{r}}}}\dd x =
{1 \over 2\ic\,\Im\pars{r}}\bracks{2\ic\,\Im\int_{0}^{1}\ln\pars{x}
\,{2r - 1 \over x - r}\,\dd x}
\\[5mm] & =
{1 \over \root{3}/2}\,\Im\pars{\bracks{-\root{3}\ic}
\int_{0}^{1}{\ln\pars{x} \over r - x}\,\dd x} =
-2\,\Re
\int_{0}^{1/r}{\ln\pars{rx} \over 1 - x}\,\dd x
\\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\,
-2\,\Re\int_{0}^{\large\bar{r}}{\ln\pars{1 - x} \over x}\,\dd x =
2\,\Re\int_{0}^{\large\bar{r}}\mrm{Li}_{2}'\pars{x}\,\dd x =
2\,\Re\mrm{Li}_{2}\pars{\exp\pars{-\,{\pi \over 3}\,\ic}}
\\[5mm] & =
2\,\Re\mrm{Li}_{2}\pars{\exp\pars{2\pi\bracks{1 \over 6}\,\ic}}
\\[5mm] & =
-\,{\pars{2\pi\ic}^{2} \over 2!}\,\
\overbrace{\mrm{B}_{2}\pars{1 \over 6}}^{\ds{1 \over 36}} =
{\pi^{2} \over 18}\qquad\qquad
\pars{~\mrm{B}_{n}\pars{x}:\ Bernoulli\ Polynomial~}
\end{align}

which is
  Junqui$\mathrm{\grave{e}}$re's Inversion Formula
  in terms of Bernoulli Polynomials $\ds{\mrm{B}_{n}\pars{x}}$. Note that $\ds{\mrm{B}_{2}\pars{x} = x^{2} - x + {1 \over 6}}$.

Then,
$$
\bbx{S \equiv
\sum_{n = 1}^{\infty}{1 \over n^{2}{2n \choose n}} = \zeta\pars{2}\,{1 \over 3}}
$$
