Advanced Integral Regarding Logarithmic Functions 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.
 A: Picking up from where you left off, let
$$I = 2 \int_0^1 \frac{1}{1 - u^2} \ln \left (\frac{1 + u}{2u} \right ) \, du.$$
Now if we enforce a substitution of 
$$x = \frac{1 + u}{2u} \quad \text{then} \quad u = \frac{1}{2x - 1},$$
and we have
$$du = -\frac{2}{(2x - 1)^2}.$$
Also note the term $1 - u^2$ under this substitution becomes
$$1 - u^2 = \frac{4x(x - 1)}{(2x - 1)^2},$$
while for the limits of integration we have: $(0,1) \mapsto (\infty,1)$. Thus
$$I = \int_1^\infty \frac{\ln x}{x(x - 1)} \, dx.$$
Next, if we enforce a substitution of $x \mapsto \frac{1}{x}$ one has
$$I = - \int^1_0 \frac{\ln (x)}{1 - x} \, dx.$$
Inside its interval of convergence the term $1/(1 - x)$ can be expanded as a geometric series. Doing so yields
$$I = - \sum_{n = 0}^\infty \int^1_0 x^n \ln (x) \, dx.$$
This last integral can be readily found using integration by parts twice. The result is
$$\int_0^1 x^n \ln (x) \, dx = -\frac{1}{(n + 1)^2}.$$
Thus
$$I = \sum_{n = 0}^\infty \frac{1}{(n + 1)^2} = \sum_{n = 1}^\infty \frac{1}{n^2} = \zeta (2),$$
where the summation index was shifted $n \mapsto n - 1$ and the well-known result for the Basel problem has been used. Thus
$$\int_0^{1/4} \frac{1}{x \sqrt{1 - 4x}} \ln \left (\frac{1 + \sqrt{1 - 4x}}{2 \sqrt{1 - 4x}} \right ) \, dx = \zeta (2) = \frac{\pi^2}{6}.$$
A: 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
$$ \mathcal{I}=4\int_{0}^{1}\frac{t}{1-t^2}\left(-\log t\right)\,dt=4\sum_{n\geq 0}\int_{0}^{1}t^{2n+1}(-\log t)\,dt=\sum_{n\geq 0}\frac{4}{(2n+2)^2}=\color{red}{\zeta(2)}. $$
