Answer
$\{x\in R: x \neq 1, x\neq -1 \}=(-\infty , -1) \cup (-1, 1) \cup (1, \infty)$
Work Step by Step
The domain of $f(x)=\frac{x+2}{x^2 -1}$ is the set of all real numbers $x$ for which $\frac{x+2}{x^2 -1}$ is defined as a real number. We know a rational expression is not defined where the denominator is zero, so we must find all values of $x$ for which $$x^2 -1 =0.$$ We add $1$ to both sides to get $$x^2=1,$$ and taking the square root of both sides gives $$x=\pm 1.$$ Thus we know $\frac{x+2}{x^2 -1}$ is not defined when $x=1$ or $x =-1$. This means the domain of $f(x)=\frac{x+2}{x^2 -1}$ is the set af all real numbers $x$ for which $x\neq 1$ and $x \neq 2$.
In set notation, the domain is $\{x\in R: x \neq 1, x\neq -1 \}.$
In interval notation, the domain is $(-\infty , -1) \cup (-1, 1) \cup (1, \infty).$
