Evaluating $\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$.

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: $$\int _0^1\frac{\ln \left(1-x^2\right)\arcsin \left(x\right)}{x^2}\:dx=-\frac{3}{2}\zeta \left(2\right)-\pi \ln \left(2\right)+4G$$

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

$$ \int_0^1{\arcsin(x)\ln(1-x^2)\over x^2}\mathrm dx=4G-{\pi^2\over4}-\pi\ln2 $$