I am not understanding those solutions that W. Mathematica gives me about this inequality:
$$\frac{x^2-1}{\vert x+1 \vert} \leq 0 $$
I solved it by hand and I find that the solution is $-1 < x \leq 1$, but Mathematica returns me $-1 \leq x \leq 1$.
I cannot consider $x = -1$, because it's not part of the domain of the funciton. It holds in "a limit sense", in my opinion.
I solved it with the absolute value definition, but there is no way I can set $x = -1$ that naively. Even if $x^2-1 = (x-1)(x+1)$ I cannot just simplify, because I would obtain a different function.
Can someone please clarify this to me?