Alternative proof of computing $\sum _{n=1}^{\infty } \frac{4^n H_n}{n^2 {2n\choose n}}$ In this solution we showed that
$$\sum _{n=1}^{\infty } \frac{4^n H_n}{n^2 {2n\choose n}}=6\ln(2)\zeta(2)+\frac72\zeta(3)\tag1$$
using the identity
$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n\choose n}}$$
My question here is can we prove $(1)$ in a different way using elementary methods? Still, don't let my question restrictions limit your approaches, all approaches are appreciated.
The point of this post (challenge) is to learn different techniques if possible and to make this site more entertaining.
Thank you.
 A: A sketch (for now). Using the identities
$$\int_0^1\frac{x^{n-1}}{\sqrt{1-x}}\,{\rm d}x=\frac{4^n}{n\binom{2n}n}\,\,\,\text{and}\,\,\,\sum_{n\ge1}\frac{H_n}nx^n=\operatorname{Li}_2(x)+\frac12\log^2(1-x)$$
gives
$$\sum_{n\ge1}\frac{4^nH_n}{n^2\binom{2n}n}=\sum_{n\ge1}\frac{H_n}n\int_0^1\frac{x^{n-1}}{\sqrt{1-x}}\,{\rm d}x=\frac12\int_0^1\frac{2\operatorname{Li}_2(x)+\log^2(1-x)}{x\sqrt{1-x}}\,{\rm d}x$$
Reflecting and enforcing $\sqrt x\mapsto x$ afterwards yields
\begin{align*}
\frac12\int_0^1\frac{2\operatorname{Li}_2(x)+\log^2(1-x)}{x\sqrt{1-x}}\,{\rm d}x&=\frac12\int_0^1\frac{2\operatorname{Li}_2(1-x)+\log^2(x)}{(1-x)\sqrt{x}}\,{\rm d}x\\
&=\int_0^1\frac{2\operatorname{Li}_2(1-x^2)+4\log^2(x)}{1-x^2}\,{\rm d}x
\end{align*}
The latter integral evaluates as $7\zeta(3)$ using the geometric series. For the first integral apply IBP two times to obtain
\begin{align*}
\int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\,{\rm d}x&=-\left[\frac12\operatorname{Li}_2(1-x^2)\log\left(\frac{1-x}{1+x}\right)\right]_0^1+2\int_0^1\frac{x\log x\log\left(\frac{1-x}{1+x}\right)}{1-x^2}\,{\rm d}x\\
&=-\left[\frac12 x\log x\log^2\left(\frac{1-x}{1+x}\right)\right]_0^1+\int_0^1(1+\log x)\log^2\left(\frac{1-x}{1+x}\right)\,{\rm d}x\\
&=\frac{\pi^2}6+\int_0^1\log x\log^2\left(\frac{1-x}{1+x}\right)\,{\rm d}x
\end{align*}
I am currently not sure how to approach the remaining integral in an elegant way.

Sidenote: Using the usual series expansion of the logarithm and the integral representation of the harmonic numbers leads to an evaluation of the remaining integral. However, this method is rather unelegant and I will see if I can find something more satisfying.
A: We use the powerful form of the Beta function presented in the book, (Almost) Impossible Integrals, Sums, and Series, $\displaystyle \int_0^1 \frac{x^{a-1}+x^{b-1}}{(1+x)^{a+b}} dx = \operatorname{B}(a,b)$, (see pages $72$-$73$).
Set $a=b=n$ we have
$$\int_0^1\frac{2x^{n-1}}{(1+x)^{2n}}dx=\frac{\Gamma^2(n)}{\Gamma(2n)}=\frac{2}{n{2n\choose n}}$$
So $$\frac{1}{n{2n\choose n}}=\int_0^1\frac{x^{n-1}}{(1+x)^{2n}}dx=\int_0^1\frac1x\left(\frac{x}{(1+x)^2}\right)^ndx$$
$$\Longrightarrow \sum_{n=1}^\infty\frac{4^nH_n}{n^2{2n\choose n}}=\int_0^1\frac1x\left(\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{4x}{(1+x)^2}\right)^n\right)dx$$
$$=\int_0^1\frac1x\left(\text{Li}_2\left(\frac{4x}{(1+x)^2}\right)+\frac12\ln\left(1-\frac{4x}{(1+x)^2}\right)\right)dx$$
$$\overset{IBP}{=}\int_0^1\frac{2+2x}{x(1-x)}\ln x\ln\left(\frac{1-x}{1+x}\right)dx$$
$$=\int_0^1\left(\frac2x+\frac{4}{1-x}\right)\ln x\ln\left(\frac{1-x}{1+x}\right)dx$$
$$\small{=2\int_0^1\frac{\ln x\ln(1-x)}{x}dx+4\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{1-x}dx}_{1-x\to x}-2\int_0^1\frac{\ln x\ln(1+x)}{x}dx-4\int_0^1\frac{\ln x\ln(1+x)}{1-x}dx}$$
$$=6\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{x}dx}_{\zeta(3)}-2\underbrace{\int_0^1\frac{\ln x\ln(1+x)}{x}dx}_{-\frac34\zeta(3)}-4\underbrace{\int_0^1\frac{\ln x\ln(1+x)}{1-x}dx}_{\zeta(3)-\frac32\ln(2)\zeta(2)}$$
$$=6\ln(2)\zeta(2)+\frac72\zeta(3)$$
The latter integral is calculated here.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{4^{n}H_{n} \over
n^{2}{2n \choose n}}} =
\int_{0}^{4}\sum_{n = 1}^{\infty}{H_{n} \over n{2n \choose n}}
\,x^{n - 1}\,\dd x
\\[5mm] = &\
\int_{0}^{4}\sum_{n = 1}^{\infty}H_{n}\, x^{n - 1}\,{\Gamma\pars{n}
\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}}\,\dd x
\\[5mm] = &\
\int_{0}^{4}\sum_{n = 1}^{\infty}H_{n}\, x^{n - 1}\int_{0}^{1}t^{n - 1}
\pars{1 - t}^{n}\,\dd t\,\dd x
\\[5mm] = &\
\int_{0}^{4}\int_{0}^{1}\sum_{n = 1}^{\infty}H_{n}
\bracks{xt\pars{1 - t}}^{\, n}\,{\dd t\,\dd x \over tx}
\\[5mm] = &\
\int_{0}^{4}\int_{0}^{1}\braces{%
-\,{\ln\pars{1 - xt\bracks{1-t}} \over 1 - xt\pars{1-t}}}
{\dd t\,\dd x \over tx}
\\[5mm] = &\
\int_{0}^{1}{2\ln^{2}\pars{\verts{1 - 2t}} +
\mrm{Li}_{2}\pars{4\bracks{1 - t})\, t}\over t}\,\dd t
\\[5mm] = &\
2\int_{-1/2}^{1/2}{2\ln^{2}\pars{\verts{2t}} +
\mrm{Li}_{2}\pars{1 - 4t^{2}}\over 1 + 2t}\,\dd t
\\[5mm] = &\
4\int_{0}^{1/2}{2\ln^{2}\pars{2t} +
\mrm{Li}_{2}\pars{1 - 4t^{2}} \over 1 - 4t^{2}}\,\dd t
\\[5mm] = &\
2\int_{0}^{1}{2\ln^{2}\pars{t} +
\mrm{Li}_{2}\pars{1 - t^{2}} \over 1 - t^{2}}\,\dd t
\\[5mm] = &\
4\
\underbrace{\int_{0}^{1}{\ln^{2}\pars{t} \over 1 - t^{2}}\,\dd t}
_{\ds{\color{red}{\LARGE\S}:\ {7 \over 4}\,\zeta\pars{3}}}\ +\
2\,
\underbrace{\int_{0}^{1}{\mrm{Li}_{2}\pars{1 - t^{2}} \over
1 - t^{2}}\,\dd t}_{\ds{\color{red}{\LARGE *}:\ {1 \over 2}\,\pi^{2}\ln\pars{2} -
{7 \over 4}\,\zeta\pars{3}}}
\\[5mm] = &\
\bbx{6\ln\pars{2}\,\zeta\pars{2} + {7 \over 2}\,\zeta\pars{3}} \\ &
\end{align}

$\left\{\begin{array}{lcl}
\ds{\color{red}{\LARGE\S}} & \ds{:} &
\mbox{First} "Partial\ Fraction\ Split.\ \mbox{Next, integrate}\ twice\ \mbox{by parts.}
\\[2mm]
\ds{\color{red}{\LARGE *}} & \ds{:} & \mbox{After integration by parts, the final expression seems to be a doable and}
\\ && \mbox{known integral.}
\end{array}\right.$
