How do I calculate this integral:$\int_{0}^{1}\ln^2 \left| \sqrt x-\sqrt{1-x} \right|dx$? Is there someone who can show me how to calculate this integral :

$\int_{0}^{1}\ln^2\left|\sqrt{x}-\sqrt{1-x}\right|dx$ ?
I substituted $x=\cos ^2 t$ and after some cleanup ended up with the expression 

$$-2 \int^{\pi/2}_{\pi/4}\log ^2(\sqrt2 \cos u)\cos (2u)du=-2 \int_0^1 \frac{\left(b^2-1\right) \log ^2(b)}{\sqrt{2-b^2}} \, db$$

Which mathematica, in its infinite wisdom, evaluates as 

$$\frac{1}{4} (4 C+2+\pi  (1+\log (2)))$$

Where C is the Catalan number.
In order to develop the integral, you should make this steps,

1 Use logarithmic properties

$\int\ln^2 \left| \sqrt x-\sqrt{1-x} \right|dx$ = $\int\ 2ln \left| \sqrt x-\sqrt{1-x} \right|dx$ = $2\int\ln \left| \sqrt x-\sqrt{1-x} \right|dx$

2 Now I would suggest you to this variable change, x=(\sin x)^2, so our new integral, using trigonometrical identities, is,
   $4\int\ln\left|(\sin x -\cos x )\right|\ sin x\ cos x dx$

3  Here use the formula, $\int u*dv$ = $u*v - \int v*du$
where u=$\int\ln(\sin x -\cos x )$
      $dv=sin x \cos x dx$,
Then, you have to solve a trigonometrical integral, which can be easily solved by the universal change of variable