#### Answer

$(-\infty, -1) \cup (-1, 1) \cup (1, +\infty) $

#### Work Step by Step

The denominator is not allowed to be equal to zero as division of zero leads to an undefined expression.
Look for the real numbers that will make the denominators of the given function equal to zero.
The first denominator, $x^2+1$, is nonzero for all real numbers. However, the second denominator is equal to zero when $x=1$ or $x=-1$.
Thus, the value of $x$ can be any real number except $1$ and $-1$.
Therefore, the domain is $(-\infty, -1) \cup (-1, 1) \cup (1, +\infty) $.