TITLE: Advanced Integral Regarding Logarithmic Functions
QUESTION [5 upvotes]: I have been trying to solve this integral to face varying success:
[The Original Problem]$$\int_{0}^{1/4}\frac{1}{x\sqrt{1-4x}}\ln\left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\right)dx$$
Although I could reduce the integral down to 
$$\int_{0}^12\frac{\ln(1+u)-\ln(2u)}{1-u^2}du$$
I could not expand the denominator in a geometric series, and I have tried some alternatives to little success.

REPLY [3 votes]: An alternative approach. The original integral equals
$$ \int_{0}^{1}\frac{1}{x\sqrt{1-x}}\log\left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)\,dx = \int_{0}^{1}\frac{2}{x\sqrt{1-x^2}}\log\left(\frac{1+\sqrt{1-x^2}}{2\sqrt{1-x^2}}\right)\,dx $$
or, by setting $x=\sin\theta$,
$$ \mathcal{I} = \int_{0}^{\pi/2}\frac{2}{\sin\theta}\log\left(\frac{1+\cos\theta}{2\cos\theta}\right)\,d\theta\stackrel{\text{IBP}}{=}-2\int_{0}^{\pi/2}\frac{\tan\frac{\theta}{2}\log\tan\frac{\theta}{2}}{\cos\theta}\,d\theta $$
such that
$$ \mathcal{I}=4\int_{0}^{1}\frac{t}{1-t^2}\left(-\log t\right)\,dt=4\sum_{n\geq 0}\int_{0}^{1}t^{2n+1}(-\log t)\,dt=\sum_{n\geq 0}\frac{4}{(2n+2)^2}=\color{red}{\zeta(2)}. $$