<s>[INST] Integration with double absolute value How to prove that?
$$
\int\left|\sqrt{|x|}\right|\ \mathrm{d}x
 = \frac{1}{3} x
   \left(\left(\sqrt{x} - \sqrt{-x}\right)\mathrm{sgn}(x)
         + \sqrt{x} + \sqrt{-x}\right)
$$
I cannot understand this result.
Where does the $1/3$ factor come out from?
 [/INST] I find the outer $| \,|$ incomprehensible since $\sqrt{|x|}\ge0$ according to the usual definition of square root.  I will answer ignoring that.
For $x\ge 0$ we have the usual result $\int x^{1/2}\,dx={2\over 3}x^{3/2}$ plus constant, which you can verify by taking the derivative of ${2\over 3}x^{3/2}$.  
Suppose $x\ge 0$; then $\mathrm{ sgn} (x)=1$, and
$$\frac{1}{3} x
   \left((\sqrt{x} - \sqrt{-x})\mathrm{sgn}(x)
         + \sqrt{x} + \sqrt{-x}\right)={1\over 3}x(2\sqrt{x})={2\over 3}x^{3/2}$$
as desired (but that raises the issue of what $\sqrt{-x}$ is, perhaps $i\sqrt{x}$).  
Now suppose $x\le 0$.  We then have $\mathrm{ sgn} (x)=-1$, and we get 
$$\frac{1}{3} x
   \left(\left(\sqrt{x} - \sqrt{-x}\right)\mathrm{sgn}(x)
         + \sqrt{x} + \sqrt{-x}\right)={1\over 3}x(2\sqrt{-x})=-{2\over 3}(-x)^{3/2}.$$
Differentiated this also gives the right answer, $\sqrt{-x}$, and thus our expression equals $\int |x|^{1/2}\,dx=\int (-x)^{1/2}\,dx$, for $x\le 0$.
</s>