An integral singular at both endpoints

Convergence. Near $0$, $\frac1{\sqrt{x(1-x)}}\sim\frac1{\sqrt x}$ and $\int_0\frac{dx}{\sqrt x}$ converges ($p=\frac12<1$). Near $1$, the integrand behaves like $\frac1{\sqrt{1-x}}$, which also gives a convergent integral. So the integral converges.

Value. Let $x=\sin^2\theta$ with $\theta\in\left(0,\frac\pi2\right)$. Then $dx=2\sin\theta\cos\theta\,d\theta$ and $\sqrt{x(1-x)}=\sin\theta\cos\theta$, so
$$\int_0^1\frac{dx}{\sqrt{x(1-x)}}=\int_0^{\pi/2}2\,d\theta=\pi.$$

Remark. Since the substitution is a bijection between the open intervals and the integrand becomes constant, the singularities disappear completely. This integral is the Beta function value $B\left(\frac12,\frac12\right)=\pi$.