Closed Form For : $$S=\sum_{k=1}^{\infty}\binom{2k}{k}^{-2}x^{k}$$
Using the Series Expansion for $\arcsin^2(x)$ one can arrive at : $$\sum_{k=0}^{\infty}\binom{2k}{k}^{-1}x^{k}=\frac{4}{4-x}-4\arcsin\left(\frac{\sqrt{x}}{2}\right)\frac{1}{\sqrt{x\left(4-x\right)}}+16\arcsin\left(\frac{\sqrt{x}}{2}\right)\frac{1}{\left(4-x\right)\sqrt{x\left(4-x\right)}}\tag{1}$$
We need to add one more Binomial Term in the Denominator and I only know of two such Integrals (though there might be others?), The Beta Integral and the Odd Powers of $\sin$.
Here I will try to use the former.
$$\frac{r}{2}\int_{0}^{1}t^{r-1}\left(1-t\right)^{r-1}dt=\binom{2r}{r}^{-1}$$
Then replacing $x\to t(1-t)x$ and Integrating from $0$ to $1$ leads to (after some more algebraic manipulations) : $$S=\frac{x}{16-x}+16\frac{x\arctan\left(\sqrt{\frac{x}{16-x}}\right)}{\left(16-x\right)\sqrt{x\left(16-x\right)}}+2x\frac{d}{dx}\sqrt{x}\int_{0}^{1}\left(\frac{\arcsin\left(\frac{\sqrt{\left(t\right)\left(1-t\right)x}}{2}\right)}{\left(4-\left(t\right)\left(1-t\right)x\right)\sqrt{\left(t\right)\left(1-t\right)\left(4-\left(t\right)\left(1-t\right)x\right)}}\right)dt$$
Though I haven't had any luck in Calculating the Integral present above.
The variation of the Integral with only $t$ instead of $t(1-t)$ can be done :
$$\int_{0}^{1}\left(\frac{\arcsin\left(\frac{\sqrt{tx}}{2}\right)}{\left(4-tx\right)\sqrt{t\left(4-tx\right)}}\right)dt=\frac{\ln\left(1-\frac{x}{4}\right)}{4\sqrt{x}}+\frac{\arcsin\left(\frac{\sqrt{x}}{2}\right)}{2\sqrt{4-x}}$$
Wolfram Alpha says "no results found in terms of standard mathematical functions".
So is it not possible to solve this integral?
Although I believe this question might have already been asked before as its the Ordinary Generating Function for Inverse Of Central Binomial Coefficient Squared but was not able to find it.