Answer
$\displaystyle \frac{4}{9}\ln|y+4|+\frac{1}{18}\ln|2y-1|+C$
Work Step by Step
Apply the partial fraction method. Factors of the denominator are linear, nonrepeated.
$\displaystyle \frac{y}{(y+4)(2y-1)}=\frac{A}{y+4}+\frac{B}{2y-1}$
$\displaystyle \frac{y}{(y+4)(2y-1)}= \displaystyle \frac{A(2y-1)+B(y+4)}{(y+4)(2y-1)}$
$y=(2A+B)y+(-A+4B)\Rightarrow\left\{\begin{array}{ll}
2A+B=1 & (I)\\
-A+4B=0 & (II)
\end{array}\right.$
$I-2\displaystyle \cdot II\Rightarrow 9B=1\Rightarrow B=\frac{1}{9}$,
Insert into $II,\ A=\displaystyle \frac{4}{9}.$
$\displaystyle \int\frac{y}{(y+4)(2y-1)}dy=\int(\frac{\frac{4}{9}}{y+4}+\frac{\frac{1}{9}}{2y-1})dy$
$= \displaystyle \frac{4}{9}\int\frac{dy}{y+4}+\frac{1}{9}\int\frac{dy}{2y-1}$
$=\displaystyle \frac{4}{9}\ln|y+4|+\frac{1}{9}\cdot\frac{1}{2}\ln|2y-1|+C$
$=\displaystyle \frac{4}{9}\ln|y+4|+\frac{1}{18}\ln|2y-1|+C$
