Answer
Converges
Work Step by Step
Since
$\int_{-1}^{1} \ln |x| d x=\int_{-1}^{0} \ln (-x) d x+\int_{0}^{1} \ln x d x $
Now, $
\int_{0}^{1} \ln x d x=\lim\limits _{b \to 0^{+}} \int_{b}^{1} \ln x d x\\
=\lim\limits _{b \to 0^{+}}[x \ln x-x]\bigg|_{b}^{1}\\
=\lim\limits _{b \to 0^{+}}[(1 (0)-1)-(b \ln b-b)]\\
=-1$
Also, $\int_{-1}^{0} \ln (-x) d x=-1 $
Then , $\int_{-1}^{1} \ln |x| d x=-1-1=-2$
Thus, the integral converges.