Chapter 4 - Inverse, Exponential, and Logarithmic Functions - Summary Exercises on Functions: Domains and Defining Equations - Exercises - Page 487: 16

$(-\infty,-\sqrt 5)\cup(-\sqrt 5,\sqrt 5)\cup(\sqrt 5,\infty)$.

The domain requirement for $f(x)=ln|x^2-5|$ is $x^2-5\ne0$ which gives $x\ne\pm\sqrt 5$ , in interval notation $(-\infty,-\sqrt 5)\cup(-\sqrt 5,\sqrt 5)\cup(\sqrt 5,\infty)$.