Answer
Converges
Work Step by Step
Since
\begin{align*}
\int_{-1}^{1}(-x \ln |x|) d x&=\int_{-1}^{0}[-x \ln (-x)] d x+\int_{0}^{1}(-x \ln x) d x\\
&=\lim _{b \rightarrow 0^{+}} \int_{-1}^{b}[-x \ln (-x)] d x+\lim _{c \rightarrow 0^{+}}\int_{c}^{1}(-x \ln x) d x \\
&=\lim _{b \rightarrow 0^{+}}\left[\frac{x^{2}}{2} \ln x-\frac{x^{2}}{4}\right]_{b}^{1}-\lim _{c \rightarrow 0^{+}}\left[\frac{x^{2}}{2} \ln x-\frac{x^{2}}{4}\right]_{c}^{1}\\
&=\lim _{b \rightarrow 0^{+}}\left[\left(\frac{1}{2} \ln 1-\frac{1}{4}\right)-\left(\frac{b^{2}}{2} \ln b-\frac{b^{2}}{4}\right)\right]-\lim _{c \rightarrow 0^{+}}\left[\left(\frac{1}{2} \ln 1-\frac{1}{4}\right)-\left(\frac{c^{2}}{2} \ln c-\frac{c^{2}}{4}\right)\right]\\
&=-\frac{1}{4}-0+\frac{1}{4}+0=0
\end{align*}
Thus the integral converges.