Show that $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n} (H_{2n} - H_n)}{4^n (2n - 1)^2} = 2 + \frac{3\pi}{2} \log(2) - 2G - \pi$$
My try : We know that $$\sum_{n=1}^{\infty} \binom{2n}{n} (H_{2n} - H_{n}) t^n = - \frac{1}{\sqrt{1 - 4t}} \log\left(\frac{1 + \sqrt{1 - 4t}}{2}\right)$$
In particular we have $$\sum_{n=1}^{\infty} \binom{2n}{n} \frac{(H_{2n} - H_{n})}{4^n} t^{2n} = - \frac{1}{\sqrt{1 - t^2}} \log\left(\frac{1 + \sqrt{1 - t^2}}{2}\right)$$....(1)
Next, dividing both sides of (1) by $t ^ 2$ and then integrating from 0 to 2, we get
$$\sum_{n=1}^{\infty} \binom{2n}{n} \frac{(H_{2n} - H_{n})}{4^n (2n - 1)} x^{2n - 1} = - \int_{0}^{x} \frac{1}{t^2 \sqrt{1 - t^2}} \log\left(\frac{1 + \sqrt{1 - t^2}}{2}\right) \, dt$$
Making the change of variable $t=sin(\theta)$ in $$J = \int \frac{1}{t^2 \sqrt{1 - t^2}} \log\left(\frac{1 + \sqrt{1 - t^2}}{2}\right) \, dt$$
$$J = \int \frac{1}{\sin^2 \theta \cos \theta} \log\left(\frac{1 + \cos \theta}{2}\right) \cos \theta \, d\theta$$