Simplification of surds $\frac{x}{\sqrt{x^2 - x^4}}$

Question

$$\frac{x}{\sqrt{x^2 - x^4}}$$

I believe that I can factor out the $x^2$ in the square root to get $$\frac{1}{\sqrt{1-x^2}} .$$

However, Wolfram Alpha doesn't do the simplification, hence my confusion.

Answer 1

Hint: $$ \sqrt{x^2}=|x| $$ and I suppose that the answer of WA comes form this

Answer 2

This is almost correct---$\sqrt{x^2}$ is $|x|$ and not $x$. We can still simplify the expression some using the identity $x = (\operatorname{sgn} x) |x|$, which gives $$\frac{x}{\sqrt{x^2 - x^4}} = \frac{(\operatorname{sgn} x) |x|}{|x| \sqrt{1 - x^2}} = \frac{\operatorname{sgn} x}{\sqrt{1 - x^2}} .$$ (Because the given expression is not defined at $x = 0$, we implicitly take the simplification not to be defined there either.)

Answer 3

$$\frac{x}{\sqrt{x^2 - x^4}}= \Bigg\{\begin{array}{c} \frac{1}{\sqrt{1-x^2}}, \ \ 0<x<1 \\ -\frac{1}{\sqrt{1-x^2}}, \ \ -1<x<0 \end{array} $$