Derive a closed form for a sum with inverse binomial coefficients First off, I would like to apologize again for the integral I posted several days ago involving $\zeta(5)$.  I was careless and did not examine the decimals out far enough.
With that said, I would now like to post a series I think is interesting. I am trying to derive a general form for 
$$ \sum_{n=1}^{\infty}\frac{nx^{n}}{\binom{2n}{n}}.$$
I thought about starting with $\displaystyle \sum_{n=1}^{\infty}\frac{2^{2n}x^{2n}}{\binom{2n}{n}}=\frac{x^{2}}{\sqrt{1-x^{2}}}+\frac{x\sin^{-1}(x)}{(1-x^{2})^{\frac{3}{2}}}$.
I tried differentiating, integrating and so forth, but it turns into a mess and I do not know how to eliminate the $2^{2n}$ nor get the $x^{2n}$ down to $x^{n}$.  Is it possible to somehow integrate in terms of, say, $t$ from $0$ to $x$? 
Any thoughts on how to go about this?.  This would then lead to $\displaystyle \sum_{n=1}^{\infty}\frac{n2^{n}}{\binom{2n}{n}}=\pi +3$ and many other forms just by using a general formula.
I ran across this in "Irresistible Integrals" by Boros and Moll. It is one of their 'Exercises'. 
Thanks very much. 
 A: Start with $$f(x) = \sum_{n=1}^{\infty}\frac{2^{2n}x^{2n}}{\binom{2n}{n}}$$ where you know that $ \displaystyle f(x) = \frac{x^{2}}{\sqrt{1-x^{2}}}+\frac{x\sin^{-1}(x)}{(1-x^{2})^{3/2}}.$ Then $$ f'(x) = \sum_{n=1}^{\infty} \frac{ 2^{2n} \cdot 2n \cdot  x^{2n-1} }{\binom{2n}{n}}.$$
Multiplying both sides by $x/2$ gives $$ \frac{x f'(x) }{2} = \sum_{n=1}^{\infty} \frac{ n 2^{2n} x^{2n} }{ \binom{2n}{n}} .$$
Then let $ \displaystyle x=\frac{ \sqrt{z} }{2} $ so 
$$ \frac{\sqrt{z} }{2\sqrt{2}} f'\left( \frac{\sqrt{z} }{2}  \right) = \sum_{n=1}^{\infty} \frac{ n z^n}{\binom{2n}{n} } .$$
Carrying out the final computation gives the sum to be
$$\frac{6z}{(z -4)^2} + \frac{ 4\sqrt{z} (z+2) \csc^{-1}(2z^{-1/2} ) }{\sqrt{4-z} (z-4)^2  } .$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{n = 1}^{\infty}{nx^{n} \over {2n \choose n}}:\ {\large ?}}$

\begin{align}&\color{#66f}{\large\sum_{n = 1}^{\infty}{nx^{n} \over {2n \choose n}}}
=\sum_{n = 1}^{\infty}nx^{n}\,
{\Gamma\pars{n + 1}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}}
=\sum_{n = 1}^{\infty}n^{2}x^{n}\,
{\Gamma\pars{n}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}}
\\[3mm]&=\sum_{n = 1}^{\infty}n^{2}x^{n}
\int_{0}^{1}t^{n - 1}\pars{1 - t}^{n}\,\dd t
=\int_{0}^{1}\sum_{n = 1}^{\infty}n^{2}\bracks{xt\pars{1 - t}}^{n}\,{\dd t \over t}
\\[3mm]&=\int_{0}^{1}
\frac{(t-1) x \left(t^2 x-t x-1\right)}{\left(t^2 x-t x+1\right)^3}\,\dd t
\\[3mm]&=\color{#66f}{\large\frac{x \left[6 \root{(4 - x) x}+4 (x+2)
\arctan\left(\root{x}/\root{4-x}\right)\right]}{(x-4)^2 \root{(4 - x) x}}}
\end{align}

