Calculate $\int_0^1\frac{(\ln x)^2}{\sqrt{x-x^2}}dx=?$ 
Find the:
$$\int_0^1\frac{(\ln x)^2}{\sqrt{x-x^2}}\,dx=\text{?}$$


My Try :
$$u=\sqrt{x-x^2} \\ du=\frac{-2x+1}{2\sqrt{x-x^2}}\,dx\\dx=\frac{2\sqrt{x-x^2}}{-2x+1}\,du$$
So we have :
$$\int_0^1\frac{(\ln x)^2}{\sqrt{x-x^2}}\,dx=\int_0^1\frac{2(\ln x)^2}{(-2x+1)}\,du$$
Now what ?
 A: This is a simple integral. By Euler's Beta function
$$ \int_{0}^{1}\frac{x^\alpha}{\sqrt{x(1-x)}}\,dx = \frac{\Gamma\left(\alpha+\tfrac{1}{2}\right)}{\Gamma(\alpha+1)}\sqrt{\pi}\tag{1}$$
for any $\alpha>-\frac{1}{2}$, hence it is enough to apply $\frac{d^2}{d\alpha^2}$ to both sides of $(1)$, then evaluate at $\alpha=0$.
It is pratical to write $\frac{dg}{d\alpha}$ as $g\cdot\frac{d}{d\alpha}\log g$ and recall that $\psi=\frac{d}{d\alpha}\log\Gamma$ and
$$ \psi(1)=-\gamma,\quad \psi\left(\tfrac{1}{2}\right)=-\gamma-\log(4),\quad \psi'(1)=\frac{\pi^2}{6},\quad \psi'\left(\tfrac{1}{2}\right)=\frac{\pi^2}{2}$$
to get
$$ \int_{0}^{1}\frac{\log^2(x)}{\sqrt{x(1-x)}}\,dx = \color{red}{\frac{\pi^3}{3}+4\pi\log^2(2)}.\tag{2}$$
This can be proved by Fourier-Legendre series expansions, too. Indeed, the hypergeometric functions mentioned by Raffaele in the comments have simple closed forms at $x\in\left\{0,\frac{1}{2},1\right\}$. Have a look at page 39 here.
Yet another (brutally efficient) approach is to apply Parseval's theorem to the Fourier series of $\log\sin$, which can be derived from the identity $\sum_{n\geq 1}\frac{\cos(n\theta)}{n}=-\log\left|2\sin\frac{\theta}{2}\right|$. This also explains why the Euler-Mascheroni constant $\gamma$ disappears from the RHS of $(2)$.
Additionally, the similar integral $\int_{0}^{1}\frac{\log^3(x)}{\sqrt{1-x^2}}\,dx$ is computed at page 76 of my notes through the same technique used above (Feynman's trick and special values for $\Gamma,\psi,\psi',\psi''$). Here there are no simple ways for applying Parseval's theorem, hence the approach by differentiation under the integral sign is a bit more general, even if Fourier series solve OP's problem sooner.
