Answer
$\frac{{10}}{9}$
Work Step by Step
$$\eqalign{
& \int_1^3 {\left( {\frac{1}{{{z^2}}} + \frac{1}{{{z^3}}}} \right)} dz \cr
& {\text{Rewrite the integrand using the property }}\frac{1}{{{x^n}}} = {x^{ - n}} \cr
& = \int_1^3 {\left( {{z^{ - 2}} + {z^{ - 3}}} \right)} dz \cr
& {\text{Integrate by the power rule }}\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr
& = \left[ {\frac{{{z^{ - 1}}}}{{ - 1}} + \frac{{{z^{ - 2}}}}{{ - 2}}} \right]_1^3 \cr
& = \left[ { - \frac{1}{z} - \frac{1}{{2{z^2}}}} \right]_1^3 \cr
& {\text{Use The Fundamental Theorem of Calculus}},{\text{ Part 2}} \cr
& = \left[ { - \frac{1}{3} - \frac{1}{{2{{\left( 3 \right)}^2}}}} \right] - \left[ { - \frac{1}{1} - \frac{1}{{2{{\left( 1 \right)}^2}}}} \right] \cr
& {\text{Simplifying}} \cr
& = - \frac{7}{{18}} + \frac{3}{2} \cr
& = \frac{{10}}{9} \cr} $$
