Answer
The integral converges to $0$.
Work Step by Step
We will compute the integral as follows:
$\int_{-1}^{1} \dfrac{dx}{x^{1/3}}= \int_{-1}^0 \dfrac{dx}{x^{1/3}}+ \int_{0}^1 \dfrac{dx}{x^{1/3}}=\lim\limits_{R \to 0^{-}} \int_{-1}^0 \dfrac{dx}{x^{1/3}}+\lim\limits_{R \to 0^{+}} \int_{R}^1 \dfrac{dx}{x^{1/3}}\\=\dfrac{3}{2} [\lim\limits_{R \to 0^{-}} [x^{2/3}]_{-1}^R+\lim\limits_{R \to 0^{+}} [x^{2/3}]_{R}^{1}]\\= \dfrac{3}{2} [\lim\limits_{R \to 0^{-}} [R^{2/3}-1]+\lim\limits_{R \to 0^{+}} (1-R^{2/3})]$
Hence, the integral converges to:
$\int_{-1}^{1} \dfrac{dx}{x^{1/3}}=\dfrac{3}{2}(-1+1)=0$
