Answer
All real numbers except $-1$ and $1$.
Work Step by Step
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 $1$.
