## Precalculus (10th Edition)

All real numbers except $-1$ and $1$.
Factor the denominator to obtain: $x^4-1=(x^2-1)(x^2+1)=(x+1)(x-1)(x^2+1)$ The denominator cannot be $0$. Hence . $(x+1)(x-1)(x^2+1)\ne0\\ \longrightarrow x+1 \ne0, x-1\ne0, \text{ and } x^2+1\ne0\\ \longrightarrow x\ne-1, x\ne1 \space \space(x^2+1\text{ has no restrictions)}$ Therefore the domain of the given function is the set of all real numbers except $-1$ and $1$.