sequences and series - Peculiar Sum regarding the Reciprocal Binomial Coefficients

Whilst playing around on Wolfram Alpha, I typed in the sum

$$\sum_{x=0}^\infty \frac{1}{\binom{2x}{x}}=\frac{2}{27}(18+\pi\sqrt 3)$$
I'm not sure how to derive the answer. My first instinct was to expand the binomial coefficient to get
$$\sum_{x=0}^\infty \frac{x!^2}{(2x)!}$$
and then to try using a Taylor Series to get the answer. I thought that if I could find a function $f(n)$ with
$$f(n)=\sum_{x=0}^\infty \frac{x!^2n^x}{(2x)!}$$
Then my sum would be equal to $f(1)$. How do I find such a function?

EDIT: I continued on this path and realized that I can use this to set up a recurrence relation for $f^{(x)}(0)$:

$$f^{(0)}(0)=1$$

$$f^{(x)}(0)=\frac{x^2}{2x(2x-1)}f^{(x-1)}(0)$$

However, I'm not sure how this helps me find $f(1)$...

Am I on the right track? Can somebody help me finish what I started, or point me towards a better method of calculating this sum?

Thanks!

Answer

Hint. One may observe that
$$\frac{1}{\binom{2n}{n}}=n\int_0^1 t^{n-1}(1-t)^ndt,\qquad n\ge1,$$ giving
$$\sum_{n=0}^\infty\frac{1}{\binom{2n}{n}}=1+\int_0^1 \sum_{n=1}^\infty nt^{n-1}(1-t)^n\:dt=1+\int_0^1\frac{t-1}{\left(t^2-t+1\right)^2}dt=\frac{2}{27} \left(18+\sqrt{3} \pi \right)$$ the latter integral is classically evaluated by partial fraction decomposition.