Answer
$$\frac{2}{3}$$
Work Step by Step
$$\eqalign{
& \int_1^3 {\frac{1}{{{x^2}}}} dx \cr
& {\text{write with negative exponent}} \cr
& = \int_1^3 {{x^{ - 2}}} dx \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \left( {\frac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}}} \right)_1^3 \cr
& = \left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right)_1^3 \cr
& = \left( { - \frac{1}{x}} \right)_1^3 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = - \frac{1}{3} - \left( { - \frac{1}{1}} \right) \cr
& {\text{simplify}} \cr
& = - \frac{1}{3} + 1 \cr
& = \frac{2}{3} \cr} $$
