A ''strange'' integral from WolframAlpha

Question

I want integrate: $$ \int \frac{1}{\sqrt{|x|}} \, dx $$ so I divide for two cases $$ x>0 \Rightarrow \int \frac{1}{\sqrt{x}} \, dx= 2\sqrt{x}+c $$

$$ x<0 \Rightarrow \int \frac{1}{\sqrt{-x}} \, dx= -2\sqrt{-x}+c $$ But WolframAlpha gives: $$ \int \frac{1}{\sqrt{|x|}} \, dx=\left(\sqrt{-x}+\sqrt{x} \right)\operatorname{sgn}(x)-\sqrt{-x}+\sqrt{x} +c $$ How I can interpret this result? Maybe I'm wrong?

Answer 1

Using $\operatorname{sgn}(x)$ is just a (half-dirty) trick to put the two cases into one. Put in $-1$ vs. $+1$ for $\operatorname{sgn}(x)$ and your eyes will be open.

Answer 2

Since the function is not defined for $x=0$, it's not really meaningful to have a single constant of integration for the whole thing. The most general function $F$ (not defined at $0$) for which, at each point $x\ne0$, $F'(x)=\frac{1}{\sqrt{|x|}}$, is $$ F(x)=\begin{cases} -2\sqrt{-x}+c_1 & \text{if $x<0$}\\ 2\sqrt{x}+c_2 & \text{if $x>0$} \end{cases} $$ where $c_1$ and $c_2$ are arbitrary constants.

Among these functions there are some that can be extended by continuity at $0$, namely those for which $c_1=c_2$, but they're just a special case. Note that none of these special functions is differentiable at $0$.