Answer
$+\infty$
The given integral is divergent.
Work Step by Step
Simplify the improper integral $\int_{-2}^{2} \dfrac{\ dx}{x^2}$
Let us consider that $I=\lim\limits_{z \to 0^{-}} \int_{-2}^{z} \dfrac{dx}{x^2}+\lim\limits_{z \to 0^{+}} \int_{z}^{2} \dfrac{dx}{x^2}$
Integrate the integral by using limits.
$I=\lim\limits_{z \to 0^{-}} [\dfrac{-1}{x}]_{-2}^a+\lim\limits_{z \to 0^{+}} [\dfrac{-1}{x}]_{z}^2 \\=(\infty-\dfrac{1}{2})+(-\dfrac{1}{2}+\infty)\\=+\infty$
So, the given integral is divergent.
