Answer
Converges
Work Step by Step
We have : $ \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\limits _{b \rightarrow 0^{+}} \int_{-1}^{b}[-x \ln (-x)] d x+\lim\limits_{c \rightarrow 0^{+}}\int_{c}^{1}(-x \ln x) d x \\
=\lim\limits _{b \rightarrow 0^{+}}[\dfrac{x^{2}}{2} \ln x-\dfrac{x^{2}}{4}]_{b}^{1}-\lim\limits_{c \rightarrow 0^{+}}[\dfrac{x^{2}}{2} \ln x-\frac{x^{2}}{4}]_{c}^{1}\\
=\lim\limits _{b \rightarrow 0^{+}}[(\dfrac{1}{2} \ln 1-\dfrac{1}{4})-t(\frac{b^{2}}{2} \ln b-\frac{b^{2}}{4})]-\lim\limits _{c \rightarrow 0^{+}}[(\dfrac{1}{2} \ln 1-\dfrac{1}{4})-(\dfrac{c^{2}}{2} \ln c-\dfrac{c^{2}}{4})]\\
=-\dfrac{1}{4}-0+\dfrac{1}{4}+0\\=0 $
Thus, the integral converges.
