Is there a closed form for $\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2?$ We have
$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot{2n-1\over 2n+1}=4-\pi\tag1$$
I would like to know if there exist a closed form for
$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2 =\,??\tag2$$
I was able to roughly estimate it as $\approx\sqrt{8+2\pi}$ but it is not the closed form.
How can we find the closd form for $(2)?$

A somewhat easier way is the following:
Let's start with 
$$\sum_{n=0}^{\infty}\frac{(2x)^{2n}}{\binom{2n}{n}}=\frac{1}{1-x^2}+\frac{x\arcsin x}{(1-x^2)^{\frac{3}{2}}}$$ 
Now, all we need is to integrate this expression twice and to differentiate it twice with respect to $x$.  
I skip these simple procedures and write down only the end result.
$$\sum_{n=0}^{\infty}\frac{(2)^{2n}(x)^{2n-2}}{\binom{2n}{n}}\left (\frac{2n-1}{2n+1}\right )^2=$$ 
$$=\frac{1}{x^2(1-x^2)}-\frac{4-5x^2}{x^3(1-x^2)^{\frac{3}{2}}}\arcsin x+\frac{2\arcsin^2 x}{x^4}+\frac{2}{x^3}\int_{0}^{x}\left (\frac{\arcsin t}{t}\right )^2dt$$  
To get the original sum, wich we denote as $S$, we evaluate this expression at $x=\frac{1}{\sqrt{2}}$ 
So the final result: 
$$S=\frac{\pi^2}{2}-3\pi+4+4\sqrt{2}\int_{0}^{\frac{1}{\sqrt{2}}}\left (\frac{\arcsin t}{t}\right )^2dt$$
