TITLE: Proving $\int_0^1\frac{\log 2-\log\left({1+\sqrt{1-x^2}}\right)}{x}dx=\frac{\left(\pi^2-12\log^22\right)}{24}$
QUESTION [18 upvotes]: $$\int_0^1\frac{\log 2-\log\left({1+\sqrt{1-x^2}}\right)}{x}dx=\frac{\left(\pi^2-12\log^22\right)}{24}$$

At first, I think it can be calculated like the following one with differential method.

$$\int_0^1\frac{x-\log\left({x+\sqrt{1-x^2}}\right)}{x}dx$$ $\\$

But, I'm wrong. I try my best to do it with differential method, but failed. Who can help me to prove it? Thank you.

REPLY [2 votes]: Let $(1+\sqrt{1-x^2})/2=t$, we get
$$\int_0^1\frac{\ln\left(\frac{1+\sqrt{1-x^2}}{2}\right)}{x}dx=\frac12\int_{1/2}^1\frac{\ln(t)}{1-t}dt-\frac12\int_{1/2}^1\frac{\ln(t)}{t}dt=-\frac14\zeta(2)+\frac12\ln^2(2).$$
A different way is by setting $\sqrt{1-x^2}=y$ then subbing $y=(1-t)/(1+t)$:
$$\int_0^1\frac{\ln\left(\frac{1+\sqrt{1-x^2}}{2}\right)}{x}dx=\int_0^1\frac{y\ln\left(\frac{1+y}{2}\right)}{1-y^2}dy$$
$$=\int_0^1\frac{\ln(1+t)}{1+t}dt-\frac12\int_0^1\frac{\ln(1+t)}{t}dt$$
$$=\frac12\ln^2(2)-\frac14\zeta(2).$$