Answer
$$\frac{1}{2}\ln 5 + \frac{1}{4}\ln 9$$
Work Step by Step
$$\eqalign{
& \int_4^8 {\frac{{ydy}}{{{y^2} - 2y - 3}}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{y}{{{y^2} - 2y - 3}} = \frac{y}{{\left( {y - 3} \right)\left( {y + 1} \right)}} \cr
& \frac{y}{{\left( {y - 3} \right)\left( {y + 1} \right)}} = \frac{A}{{y - 3}} + \frac{B}{{y + 1}} \cr
& {\text{Multiplying }}\left( {y - 3} \right)\left( {y + 1} \right){\text{, we have}} \cr
& y = A\left( {y + 1} \right) + B\left( {y - 3} \right) \cr
& \cr
& {\text{if we set }}y = 3 \cr
& 3 = A\left( {3 + 1} \right) + B\left( {3 - 3} \right) \cr
& 3 = A\left( 4 \right) \cr
& A = \frac{3}{4} \cr
& \cr
& {\text{if we set }}y = - 1 \cr
& - 1 = A\left( { - 1 + 1} \right) + B\left( { - 1 - 3} \right) \cr
& - 1 = B\left( { - 4} \right) \cr
& B = \frac{1}{4} \cr
& \cr
& {\text{then}} \cr
& \frac{y}{{\left( {y - 3} \right)\left( {y + 1} \right)}} = \frac{{3/4}}{{y - 3}} + \frac{{1/4}}{{y + 1}} \cr
& \int_4^8 {\frac{{ydy}}{{{y^2} - 2y - 3}}} = \int_4^8 {\left( {\frac{{3/4}}{{y - 3}} + \frac{{1/4}}{{y + 1}}} \right)} dy \cr
& {\text{integrating}} \cr
& = \left( {\frac{3}{4}\ln \left| {y - 3} \right| + \frac{1}{4}\ln \left| {y + 1} \right|} \right)_4^8 \cr
& = \left( {\frac{3}{4}\ln \left| {8 - 3} \right| + \frac{1}{4}\ln \left| {8 + 1} \right|} \right) - \left( {\frac{3}{4}\ln \left| {4 - 3} \right| + \frac{1}{4}\ln \left| {4 + 1} \right|} \right) \cr
& {\text{simplifying}} \cr
& = \frac{3}{4}\ln 5 + \frac{1}{4}\ln 9 - \frac{3}{4}\ln 1 - \frac{1}{4}\ln 5 \cr
& = \frac{1}{2}\ln 5 + \frac{1}{4}\ln 9 \cr} $$
