A Challenging Euler Sum $\sum\limits_{n=1}^\infty \frac{H_n}{\tbinom{2n}{n}}$ Recently, I encountered a strange series involving Harmonic Numbers and Binomial Coefficients both.
According to Mathematica:
$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}} = -\frac{2\sqrt{3} \pi}{27}(\log (3)-2)+\frac{2}{27} \left( \psi_1 \left( \frac{1}{3}\right)-\psi_1 \left(\frac{2}{3} \right)\right)$$
Here $\psi_n(z)$ denotes the Polygamma Function. Can anybody provide a nice proof of the above statement? 
My Failed Attempt
Using the Beta-function identity,
$$\frac{1}{\binom{2n}{n}}=(2n+1)\int_0^1 y^n(1-y)^n \ dy$$
$$\displaystyle \begin{aligned} \sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}} &= \sum_{n=1}^\infty (2n+1)H_n \int_0^1 (y-y^2)^n dy \\  &= \int_0^1 \sum_{n=1}^\infty (2n+1)H_n (y-y^2)^n \ dy \end{aligned}$$
Here, I used the identity
$$\sum_{n=1}^\infty (2n+1)H_n t^n=\frac{2t-(1+t)\log(1-t)}{(t-1)^2}\quad |t|<1$$ 
and got
$$\sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}}=\int_0^1 \frac{2y-2y^2-(1+y-y^2)\log(y^2-y+1)}{(y^2-y+1)^2}dy$$
How should I continue from here? I tried making some substitutions but nothing worked. Am I going in the right direction?
Please help.
 A: Note: there is a minus sign missing in front of the 1st term on the left side of your evaluation.

Put the integral into the form
$$I=\int_0^{1}\frac{\left(y^2-y+1\right)\ln\left(y^2-y+1\right)-2\ln\left(y^2-y+1\right)-2\left(y^2-y+1\right)+2}{\left(y^2-y+1\right)^2}dy.$$
Making the change of variable $y=\frac12+\frac{\sqrt{3}}{2}\tan\varphi$ (so that $y^2-y+1=\frac{3}{4\cos^2\varphi}$) and simplifying, this reduces to
$$I=\frac{8}{3\sqrt{3}}\int_{-\pi/6}^{\pi/6}\left\{\Bigl(\frac34-2\cos^2 \varphi\right)\left(\ln 3-2-2\ln (2\cos \varphi)\Bigr)-2\cos^2 \varphi\right\}d\varphi.\tag{1}$$
The only nontrivial integrals here are of the form
$$\int_{-\pi/6}^{\pi/6}\ln (2\cos \varphi)\,d\varphi,\qquad \int_{-\pi/6}^{\pi/6}\left(2\cos^2\varphi-1\right)\ln (2\cos \varphi)\,d\varphi.$$
The second integral can be easily done by parts - it is equal to 
\begin{align}\int_{-\pi/6}^{\pi/6}\left(2\cos^2\varphi-1\right)\ln (2\cos \varphi)\,d\varphi&=\Bigl[\sin\varphi\cos\varphi\ln(2\cos \varphi)\Bigr]^{\pi/6}_{-\pi/6}+\int_{-\pi/6}^{\pi/6}\sin^2\varphi\,d\varphi=\\&=\frac{\pi}{6}+\frac{\sqrt{3}}{4}\ln3-\frac{\sqrt{3}}{4}.
\end{align}
Using this in (1), we reduce it to
$$I=\frac{8}{3\sqrt{3}}\left[\frac{\pi\left(2-\ln3\right)}{12}+\frac12\int_{-\pi/6}^{\pi/6}\ln(2\cos\varphi)\,d\varphi\right].$$
Therefore, the proof of your identity reduces to showing that
$$\int_{0}^{\pi/6}\ln(2\cos\varphi)\,d\varphi=\frac{\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)}{12\sqrt{3}},\tag{2}$$
However, the left side is clearly expressible in terms of polylogarithms, so (2) should follow from their known special values.

Added: 
Indeed, as Raymond Manzoni pointed out, the difference of the  formulas (5) and (7) here gives
$$\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)=6\sqrt{3}\,\mathrm{Cl}_2\left(\frac{2\pi}{3}\right)\tag{3}$$
Clausen function $\mathrm{Cl}_2\left(x\right)$ is basically the imaginary part of dilogarithm function, characterized by the integral representation
$$\mathrm{Cl}_2\left(x\right)=-\int_0^x\ln\left(2\sin\frac{t}{2}\right)dt.\tag{4}$$
Using (3), (4) and the fact that $\mathrm{Cl}_2(\pi)=0$, we deduce from (2) the necessary statement.
A: You can refer to my calculation in this forum.
