Answer
$${\text{the improper integral diverges}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{1}{{{x^2}}}} dx \cr
& \frac{1}{{{x^2}}}{\text{ has an infinite discontinuity at }}x = 0,{\text{ so we can write}} \cr
& \int_0^1 {\frac{1}{{{x^2}}}} dx = \mathop {\lim }\limits_{b \to {0^ + }} \int_b^1 {\frac{1}{{{x^2}}}} dx \cr
& {\text{Integrate}} \cr
& = \mathop {\lim }\limits_{b \to {0^ + }} \left[ { - \frac{1}{x}} \right]_b^1 \cr
& = \mathop {\lim }\limits_{b \to {0^ + }} \left[ { - \frac{1}{1} - \frac{1}{b}} \right] \cr
& = \mathop {\lim }\limits_{b \to {0^ + }} \left[ { - 1 + \frac{1}{b}} \right] \cr
& {\text{Evaluate the limit}} \cr
& = - 1 + \frac{1}{{{0^ + }}} \cr
& = - 1 + \infty \cr
& = \infty \cr
& {\text{So}},{\text{ the improper integral diverges}}. \cr
& {\text{The following graph confirms the result}} \cr} $$
