Evaluating $\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$. My attempt.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$$
$$=\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k}\binom{2k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}$$
The first sum can be evaluated easily if one uses the central binomial coefficient generating function
, the closed form is $2\zeta \left(2\right)$.

For the remaining sum consider the $\arcsin$ series expansion.
$$\sum _{k=0}^{\infty }\frac{x^{2k+1}}{4^k\left(2k+1\right)}\binom{2k}{k}=\arcsin \left(x\right)$$
$$\sum _{k=1}^{\infty }\frac{x^k}{4^k\left(2k+1\right)}\binom{2k}{k}=\frac{\arcsin \left(\sqrt{x}\right)}{\sqrt{x}}-1$$
$$-\sum _{k=1}^{\infty }\frac{1}{4^k\left(2k+1\right)}\binom{2k}{k}\int _0^1x^{k-1}\ln \left(1-x\right)\:dx=-\int _0^1\frac{\arcsin \left(\sqrt{x}\right)\ln \left(1-x\right)}{x\sqrt{x}}\:dx$$
$$+\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx$$
$$\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}=-2\int _0^1\frac{\arcsin \left(x\right)\ln \left(1-x^2\right)}{x^2}\:dx-\zeta \left(2\right)$$
But I got stuck with:
$$\int _0^1\frac{\arcsin \left(x\right)\ln \left(1-x^2\right)}{x^2}\:dx$$
Anything I try yields more complicated stuff, is there a way to calculate the main sum or the second one (or the integral) elegantly\in a simple manner?
 A: It seemed something was missing, so with the right tools the proof isn't difficult.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$$

Consider:
$$\sum _{k=1}^{\infty }\frac{x^k}{4^k}H_k\binom{2k}{k}=\frac{2}{\sqrt{1-x}}\ln \left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)$$
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k}\binom{2k}{k}\int _0^1x^{2k}\:dx=2\int _0^1\frac{\ln \left(1+\sqrt{1-x^2}\right)}{\sqrt{1-x^2}}\:dx-2\int _0^1\frac{\ln \left(\sqrt{1-x^2}\right)}{\sqrt{1-x^2}}\:dx$$
$$-2\ln \left(2\right)\int _0^1\frac{1}{\sqrt{1-x^2}}\:dx$$
$$=2\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx-2\int _0^1\frac{\ln \left(x\right)}{\sqrt{1-x^2}}\:dx-\pi \ln \left(2\right)$$

$$\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx=\frac{\pi }{2}\ln \left(2\right)-\int _0^1\frac{\arcsin \left(x\right)}{1+x}\:dx$$
$$=\frac{\pi }{2}\ln \left(2\right)-\int _0^{\frac{\pi }{2}}\frac{x\cos \left(x\right)}{1+\sin \left(x\right)}\:dx=\int _0^{\frac{\pi }{2}}\ln \left(1+\sin \left(x\right)\right)\:dx$$
$$=4\int _0^1\frac{\ln \left(1+t\right)}{1+t^2}\:dt-2\int _0^1\frac{\ln \left(1+t^2\right)}{1+t^2}\:dt$$
This means that:
$$\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx=-\frac{\pi }{2}\ln \left(2\right)+2G$$

Thus:
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}=-\pi \ln \left(2\right)+4G$$

Bonus.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}=\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k}\binom{2k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}$$
And so we find that:
$$\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}=2\zeta \left(2\right)+2\pi \ln \left(2\right)-8G$$
And in the body of the question we had:
$$\int _0^1\frac{\ln \left(1-x^2\right)\arcsin \left(x\right)}{x^2}\:dx=-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}-\frac{1}{2}\zeta \left(2\right)$$
Hence:
$$\int _0^1\frac{\ln \left(1-x^2\right)\arcsin \left(x\right)}{x^2}\:dx=-\frac{3}{2}\zeta \left(2\right)-\pi \ln \left(2\right)+4G$$
A: Since $x\in(0,1)$, we can utilize the power series for natural log:
$$
\begin{aligned}
I
&=\int_0^1{\arcsin(x)\ln(1-x^2)\over x^2}\mathrm dx \\
&=-\int_0^1\arcsin(x)\sum_{k=1}^\infty{x^{2k-2}\over k}\mathrm dx \\
&=-\sum_{k=1}^\infty\frac1k\underbrace{\int_0^1x^{2k-2}\arcsin(x)\mathrm dx}_{x=\sin\theta} \\
&=-\sum_{k=1}^\infty\frac1k\left[\left.{x^{2k-1}\arcsin(x)\over2k-1}\right|_0^1-{1\over2k-1}\underbrace{\int_0^1{x^{2k-1}\over\sqrt{1-x^2}}\mathrm dx}_{x=\sqrt t}\right] \\
&=-\sum_{k=1}^\infty\frac1k\left[{\pi\over2(2k-1)}-{1\over2(2k-1)}\int_0^1t^{k-1}(1-t)^{1/2-1}\mathrm dt\right] \\
&=-\sum_{k=1}^\infty{1\over2k(2k-1)}\left[\pi-B\left(k,\frac12\right)\right] \\
&=\sum_{k=1}^\infty{1\over2k(2k-1)}{\Gamma(k)\Gamma\left(\frac12\right)\over\Gamma\left(k+\frac12\right)}-\pi\sum_{k=1}^\infty{1\over2k(2k-1)}
\end{aligned}
$$
For the last term, we have
$$
\sum_{k=1}^\infty{1\over2k(2k-1)}=\sum_{k=1}^\infty\left[{(-1)^{2k-1+1}\over2k-1}+{(-1)^{2k+1}\over2k}\right]=\sum_{n=1}^\infty{(-1)^{n+1}\over n}=\ln2
$$
For the first term, by Legendre's duplication formula we have
$$
\Gamma\left(k+\frac12\right)=2^{1-2k}\sqrt\pi{\Gamma(2k)\over\Gamma(k)}
$$
which leads to
$$
\begin{aligned}
\sum_{k=1}^\infty{1\over2k(2k-1)}{\Gamma(k)\Gamma\left(\frac12\right)\over\Gamma\left(k+\frac12\right)}
&=\frac12\sum_{k=1}^\infty{4^k\over2k-1}{[(k-1)!]^2\over(2k)!} \\
&=\frac12\sum_{k=1}^\infty{[(k-1)!]^2\over(2k)!}\int_0^4t^{2k-2}\mathrm dt
\end{aligned}
$$
Due to lack of necessary skills, I am not able to continue from this point, but Mathematica gives $4G-{\pi^2\over4}$ where $G$ is Catalan's constant:
$$
G\triangleq\sum_{n=0}^\infty{(-1)^n\over(2n+1)^2}
$$
As a result, the integral evaluates to
$$
\int_0^1{\arcsin(x)\ln(1-x^2)\over x^2}\mathrm dx=4G-{\pi^2\over4}-\pi\ln2
$$
