Answer
$= \ln \frac{1}{3}=-\ln{3}$
Work Step by Step
$\int^{3}_1 \frac{y^{3}-2y^{2}-y}{y^{2}} dy$
$= \int^{3}_1 \frac{y^{3}}{y^{2}}-\frac{2y^{2}}{y^{2}}-\frac{y}{y^{2}}dy$
$= \int^{3}_1 y-2-\frac{1}{y}dy$
$= \int^{3}_1 y-2-y^{-1}dy$
$= \frac{y^{2}}{2} - 2y - \ln |y| |^{3}_1$
$= [\frac{3^{2}}{2} - 2(3) - \ln |3|] - [\frac{1^{2}}{2} - 2(1) - \ln |1|]$
$= [\frac{9}{2} - 6 - \ln 3] - [\frac{1}{2} - 2 - \ln 1]$
$= [-\frac{3}{2} - \ln 3] - [-\frac{3}{2} - \ln 1]$
$= -\frac{3}{2} - \ln 3 +\frac{3}{2} + \ln 1$
$= - \ln 3 + \ln 1$
$= \ln 1 - \ln 3$
$= \ln \frac{1}{3}$
