<s>[INST] 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?
 [/INST] 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 $\pi^2$.
</s>