Answer
$- \ln 3$
Work Step by Step
$$\eqalign{
& \int_1^3 {\frac{{{y^3} - 2{y^2} - y}}{{{y^2}}}} dy \cr
& {\text{Distribute the numerator}} \cr
& = \int_1^3 {\left( {\frac{{{y^3}}}{{{y^2}}} - \frac{{2{y^2}}}{{{y^2}}} - \frac{y}{{{y^2}}}} \right)} dy \cr
& {\text{Simplify}} \cr
& = \int_1^3 {\left( {y - 2 - \frac{1}{y}} \right)} dy \cr
& {\text{Integrate by using: }}\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + C{\text{ and }}\int {\frac{1}{x}} dx = \ln \left| x \right| + C \cr
& = \left[ {\frac{1}{2}{y^2} - 2y - \ln \left| y \right|} \right]_1^3 \cr
& {\text{Use The Fundamental Theorem of Calculus}},{\text{ Part 2}} \cr
& = \left[ {\frac{1}{2}{{\left( 3 \right)}^2} - 2\left( 3 \right) - \ln \left| 3 \right|} \right] - \left[ {\frac{1}{2}{{\left( 1 \right)}^2} - 2\left( 1 \right) - \ln \left| 1 \right|} \right] \cr
& {\text{Simplifying}} \cr
& = \left[ {\frac{9}{2} - 6 - \ln 3} \right] - \left[ {\frac{1}{2} - 2 - \ln \left| 1 \right|} \right] \cr
& = - \frac{3}{2} - \ln 3 + \frac{3}{2} \cr
& = - \ln 3 \cr} $$
