Answer
$$\frac{1}{3}$$
Work Step by Step
$$\eqalign{
& \int_3^\infty {\frac{1}{{{x^2}}}} dx \cr
& {\text{solve the improper integral using the definition }}\int_a^\infty {f\left( x \right)} dx = \mathop {\lim }\limits_{b \to \infty } \int_a^b {f\left( x \right)} dx{\text{ }} \cr
& {\text{then}} \cr
& \int_3^\infty {\frac{1}{{{x^2}}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_3^b {\frac{1}{{{x^2}}}} dx{\text{ }} \cr
& {\text{write }}\frac{1}{{{x^2}}}{\text{ as }}{x^{ - 2}} \cr
& \int_3^\infty {\frac{1}{{{x^2}}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_3^b {{x^{ - 2}}} dx{\text{ }} \cr
& {\text{integrate by using the power property }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr
& \int_3^\infty {\frac{1}{{{x^2}}}} dx = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right)_3^b \cr
& \int_3^\infty {\frac{1}{{{x^2}}}} dx = - \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{x}} \right)_3^b \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& \int_3^\infty {\frac{1}{{{x^2}}}} dx = - \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{b} - \frac{1}{3}} \right) \cr
& {\text{evaluate the limit when }}b \to \infty \cr
& \int_3^\infty {\frac{1}{{{x^2}}}} dx = - \left( {\frac{1}{\infty } - \frac{1}{3}} \right) \cr
& {\text{Simplify}} \cr
& \int_3^\infty {\frac{1}{{{x^2}}}} dx = - \left( { - \frac{1}{3}} \right) \cr
& \int_3^\infty {\frac{1}{{{x^2}}}} dx = \frac{1}{3} \cr} $$