Answer
$-2$, converges
Work Step by Step
Since
$$\int_{-1}^{1} \ln |x| d x=\int_{-1}^{0} \ln (-x) d x+\int_{0}^{1} \ln x d x $$
and
\begin{align*}
\int_{0}^{1} \ln x d x&=\lim _{b \rightarrow 0^{+}} \int_{b}^{1} \ln x d x\\
&=\lim _{b \rightarrow 0^{+}}[x \ln x-x]\bigg|_{b}^{1}\\
&=\lim _{b \rightarrow 0^{+}}[(1 \cdot 0-1)-(b \ln b-b)]\\
&=-1
\end{align*}
also,
$$\int_{-1}^{0} \ln (-x) d x=-1 $$
Then
$$\int_{-1}^{1} \ln |x| d x=-2$$
