Answer
The integral was evaluated as if the integrand is continuous on the interval.
It is not, so the integral is improper, and diverges.
Using a CAS, the result is $\infty$ (diverges).
Work Step by Step
$\displaystyle \int_{-1}^{1}\frac{1}{x^{2}}dx$ is an improper integral, because $\displaystyle \frac{1}{x^{2}}$ is undefined at $x=0\in[-1,1].$
From the Definition of Improper Integrals with Infinite Discontinuities,
$\displaystyle \int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx$
$\displaystyle =\lim_{c\rightarrow b^{-}}\int_{a}^{c}f(x)dx+\lim_{c\rightarrow a^{+}}\int_{c}^{b}f(x)dx$.
$\displaystyle \int_{-1}^{1}\frac{1}{x^{2}}dx=\lim_{c\rightarrow 0^{-}}\int_{-1}^{c}x^{-2}dx+\lim_{c\rightarrow 0^{+}}\int_{c}^{1}x^{-2}dx$.
$=\displaystyle \lim_{c\rightarrow 0^{-}}[-\frac{1}{x}]_{-1}^{0}+\lim_{c\rightarrow 0^{+}}[-\frac{1}{x}]_{0}^{1}$
$=(\infty+1)+(-1+\infty)$
$=\infty$
Diverges.
The integral was evaluated as if the integrand is continuous on the interval.
It is not, so the integral is improper, and diverges.
Using a CAS, the result is $\infty$ (diverges).
