Showing that $\sum_{n=0}^{\infty}\frac{2^{n+2}}{{2n\choose n}}\cdot\frac{n-1}{n+1}=(\pi-2)(\pi-4)$ Showing that
(1)
$$\sum_{n=0}^{\infty}\frac{2^{n+2}}{{2n\choose n}}\cdot\frac{n-1}{n+1}=(\pi-2)(\pi-4)$$

see here
(2)
$$(\arcsin(x))^2=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
Look very much similar to (1). How can I make the use of (2) to solve (1)?
Let try and differentiate (1)
$$\frac{d}{dx}\arcsin(x))^2=\frac{d}{dx}\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
$$\arcsin(x)\cdot\frac{1}{\sqrt{1-x^2}}=2\sum_{n=1}^{\infty}\frac{(2x)^{2n-1}}{n{2n\choose n}}$$
Getting close but not near yet
I am stuck, any help please?
 A: Hint. By using the Euler beta function, one has
$$
\frac{2^{n+2}(n-1)}{{2n \choose n}(n+1)}=\frac{4(n-1)(2n+1)}{n+1}\int_0^1(2x(1-x))^ndx,\quad n \geq 0. \tag1
$$ Then, one may obtain
$$\begin{align}
&\sum_0^\infty\frac{2^{n+2}(n-1)}{{2n \choose n}(n+1)}
\\\\&=\sum_0^\infty\frac{4(n-1)(2n+1)}{n+1}\int_0^1(2x(1-x))^ndx\\\\
&=\int_0^1\sum_0^\infty\frac{4(n-1)(2n+1)}{n+1}(2x(1-x))^ndx \:dx\\\\
&=\int_0^1\left(\frac{-40x^2+40x-12}{(1-2 x+2 x^2)^2}-\frac{4 \log(1-2 x+2 x^2)}{(1-x) x}\right)dx\\\\
&=\underbrace{\int_0^1\frac{-40x^2+40x-12}{(1-2 x+2 x^2)^2}\,dx}_{\large \color{blue}{8-6\pi}}+\underbrace{4\int_0^1-\frac{\log(1-2 x+2 x^2)}{(1-x) x}\,dx}_{\large \color{red}{\pi^2}}
\\\\&=\color{blue}{8-6\pi}+\color{red}{\pi^2}
\\\\&=(\pi-2)(\pi-4)
\end{align}
$$ 
as announced.

Some details. 
Observe that
$$
\sum_{n=0}^\infty(2n-3)y^n=\frac{5 y-3}{(1-y)^2},\quad |y|<1, \tag{A}
$$
$$
\sum_{n=0}^\infty\frac1{n+1}y^n=-\frac{\log(1-y)}y,\quad |y|<1, \tag{B}
$$ One has
$$
\frac{4(n-1)(2n+1)}{n+1}=4(2 n-3)+\frac8{1+n}  \tag{C}
$$ then, using $(\text{A})$, $(\text{B})$ and $(\text{C})$ with $y:=2x(1-x)$ gives
$$
\sum_0^\infty\frac{4(n-1)(2n+1)}{n+1}(2x(1-x))^n=\frac{-40x^2+40x-12}{(1-2 x+2 x^2)^2}-\frac{4 \log(1-2 x+2 x^2)}{(1-x) x}
$$ One may classically integrate the fraction by parts. One may then rewrite the  $\log$-term as
$$
-\frac{4 \log(1-2 x+2 x^2)}{(1-x) x}=\frac{16}{1-u^2}\log \left(1-\frac{1-u^2}2\right) \tag{D}
$$ with $u:=2x-1$, expanding in power series of $(1-u^2)$, integrating termwise to recognize a classic binomial series giving $\pi^2$.
A: It is worth considering that:
$$ \frac{1}{\binom{2n}{n}}=\frac{n!^2}{(2n)!}=(2n+1)\cdot B(n+1,n+1)=(2n+1)\int_{0}^{1}x^n(1-x)^n\,dx \tag{1}$$
so:
$$\sum_{n\geq 0}\frac{2^{n+2}}{\binom{2n}{n}}=\int_{0}^{1}\frac{4+8x(1-x)}{(1-2x(1-x))^2}\,dx = 4\int_{-1}^{1}\frac{3-y^2}{(1+y^2)^2}\,dy=8+2\pi\tag{2}$$
as well as:
$$\begin{eqnarray*} \sum_{n\geq 0}\frac{2^{n+2}}{(n+1)\binom{2n}{n}}&=&4\int_{0}^{1}\int_{0}^{1}\frac{1+2xy-2x^2y}{(1-2xy+2x^2 y)^2}\,dx\,dy \\&=&8\int_{0}^{1}\frac{dx}{1-2x(1-x)}+2\int_{0}^{1}\frac{\log(1-2x(1-x))}{x(1-x)}\tag{3}\end{eqnarray*}$$
that is easy to compute through the substitution $y=2x-1$ as in $(2)$.
A: I have added a hand written image of the solution.  The place where you need to take the results as is the sum of reciprocals of central binomial coefficients where people have given you articles to seive through.  Goodluck.

