Answer
$(-\infty,\ -1)\cup (-1,1)\cup (1,\ \infty)$
Work Step by Step
We are given:
$f(x)= \frac{x+2}{x^{2}-1}$
The domain of a function consists of all the values that $x$ is allowed to have. In this case, we can not have division by $0$:
$x^{2}-1\neq 0$
$ x^{2}\neq 1$
$ x\neq\pm 1$
Therefore, the domain is: $(-\infty,\ -1)\cup (-1,1)\cup (1,\ \infty)$
