Answer
$\displaystyle \frac{4}{9}\ln|y+4|+\frac{1}{18}\ln|2y-1|+C$
Work Step by Step
Partial Fraction Decomposition: $\displaystyle \quad\frac{px+q}{(x-a)(x-b)} = \frac{A}{x-a}+\frac{B}{x-b}$
$\displaystyle \int\frac{y}{(2x-1)(x+4)}dy$
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Partial Fraction Decomposition:
$y=A(y+4)+B(2y-1)$
Let $y = \displaystyle -4\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$Let $\displaystyle y=\frac{1}{2}$
$\displaystyle -4=B(2(-4)-1))\quad\quad\quad\quad\quad\quad \frac{1}{2}=A(\frac{1}{2}+4)$
$\displaystyle -4=-9B\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$ $ \displaystyle \frac{1}{2}=\frac{9}{2}A$
$\displaystyle \frac{4}{9}=B\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$ $\displaystyle \frac{1}{9}=A$
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$\displaystyle \int\frac{y}{(2x-1)(x+4)}dy = \int\frac{4/9}{y+4}+\frac{1/9}{2y-1}dy=\int\frac{4/9}{y+4}dy+\int\frac{1/9}{2y-1}dy$
$\displaystyle \frac{4}{9}\ln|y+4|+\frac{1}{9}\int\frac{1}{2y-1}dy\qquad$ Let $u=2y-1\quad \rightarrow \quad du=2dy$
$\displaystyle \frac{4}{9}\ln|y+4|+\frac{1}{2}\cdot\frac{1}{9}\int\frac{1}{u}du$
$\displaystyle \frac{4}{9}\ln|y+4|+\frac{1}{18}\ln|u|+C$
$\displaystyle \frac{4}{9}\ln|y+4|+\frac{1}{18}\ln|2y-1|+C$
