jmc

The expression is defined as long as the denominator $|x-3|+|x+1|$ is not equal to 0. Since the absolute value function is always non-negative, the only way that $|x-3|+|x+1|=0$ is if both $|x-3|$ and $|x+1|$ are equal to 0. In turn, this occurs if and only if $x=3$ and $x=-1$. Clearly, $x$ cannot be both 3 and $-1$ at the same time, so the denominator is always non-zero. Therefore, the domain of the function is $\boxed{(-\infty ,\infty )}.$