Answer
$$\ln 2 + 3ln\frac{3}{2}=\ln{\frac{27}{4}}$$
Work Step by Step
$$\eqalign{
& \int_{1/2}^1 {\frac{{y + 4}}{{{y^2} + y}}} dy \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{{y + 4}}{{{y^2} + y}} = \frac{{y + 4}}{{y\left( {y + 1} \right)}} \cr
& \frac{{y + 4}}{{y\left( {y + 1} \right)}} = \frac{A}{y} + \frac{B}{{y + 1}} \cr
& {\text{Multiplying }}y\left( {y + 1} \right){\text{, we have}} \cr
& y + 4 = A\left( {y + 1} \right) + By \cr
& \cr
& {\text{if we set }}y = 0 \cr
& 0 + 4 = A\left( {0 + 1} \right) + B\left( 0 \right) \cr
& 4 = A\left( 1 \right) \cr
& A = 4 \cr
& \cr
& {\text{if we set }}y = - 1 \cr
& - 1 + 4 = A\left( { - 1 + 1} \right) + B\left( { - 1} \right) \cr
& 3 = B\left( { - 1} \right) \cr
& B = - 3 \cr
& \cr
& {\text{then}} \cr
& \frac{{y + 4}}{{y\left( {y + 1} \right)}} = \frac{4}{y} + \frac{{ - 3}}{{y + 1}} \cr
& \cr
& \int_{1/2}^1 {\frac{{y + 4}}{{{y^2} + y}}} dy = \int_{1/2}^1 {\left( {\frac{4}{y} - \frac{3}{{y + 1}}} \right)} dy \cr
& {\text{integrating}} \cr
& = \left( {4\ln \left| y \right| - 3\ln \left| {y + 1} \right|} \right)_{1/2}^1 \cr
& = \left( {4\ln \left| 1 \right| - 3\ln \left| {1 + 1} \right|} \right) - \left( {4\ln \left| {\frac{1}{2}} \right| - 3\ln \left| {\frac{1}{2} + 1} \right|} \right) \cr
& {\text{simplifying}} \cr
& = - 3\ln 2 + 4\ln 2 + 3ln\frac{3}{2} \cr
& = \ln 2 + 3ln\frac{3}{2} \cr} $$
