Answer
$\frac{9}{5}\ln{\frac{8}{3}}$
Work Step by Step
$\int_1^2{\frac{4y^{2}-7y-12}{y(y+2)(y-3)}}dy$ = $\int_1^2\left({\frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-3}}\right)dy$
$4y^{2}-7y-12$ = $A(y+2)(y-3)+By(y-3)+Cy(y+2)$
$4y^{2}-7y-12$ = $A(y^{2}-y-6)+B(y^{2}-3y)+C(y^{2}+2y)$
$$\begin{cases}
A+B+C& = 4\\
-A-3B+2C&= -7\\
-6A& = -12.
\end{cases}$$
$A = 2$, $B = \frac{9}{5}$, $C = \frac{1}{5}$
So
$\int_1^2{\frac{4y^{2}-7y-12}{y(y+2)(y-3)}}dy$ = $\int_1^2\left[{\frac{2}{y}+\frac{9}{5(y+2)}+\frac{1}{5(y-3)}}\right]dy$ = $\left[2\ln{y}+\frac{9}{5}{\ln|y+2|}+\frac{1}{5}{\ln|y-3|}\right]_1^2$ = $(2\ln2+\frac{9}{5}\ln4+\frac{1}{5}\ln1)-(2\ln1+\frac{9}{5}\ln3+\frac{1}{5}\ln2)$ = $\frac{9}{5}\ln{\frac{8}{3}}$