Answer
$\displaystyle \frac{27\ln 2-9\ln 3}{5} \approx 1.76549$
Work Step by Step
$\displaystyle \frac{4y^{2}-7y-12}{y(y+2)(y-3)}=\frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-3}$
$4y^{2}-7y-12=A(y+2)(y-3)+By(y-3)+Cy(y+2)$
$4y^{2}-7y-12=A(y^{2}-y-6)+By^{2}-3By+Cy^{2}+2Cy$
$4y^{2}-7y-12=Ay^{2}-Ay-6A+By^{2}-3By+Cy^{2}+2Cy$
$4y^{2}-7y-12=(A+B+C)y^{2}+(-A-3B+2C)y+(-6A)$
$\left\{\begin{array}{llllll}
-6A=-12 & \Rightarrow & A & & =2 & \\
-A-3B+2C=-7 & \Rightarrow & -3B & +2C & =-5 & ...I\\
A+B+C=4 & \Rightarrow & B & +C & =2 & ...II
\end{array}\right.$
$I+3(II)\displaystyle \Rightarrow 5C=1\Rightarrow C=\frac{1}{5}$
$II\displaystyle \Rightarrow B=2-\frac{1}{5}=\frac{9}{5}$
$\displaystyle \int_{1}^{2}\dfrac{4y^{2}-7y-12}{y(y+2)(y-3)}dy=\int_{1}^{2}(\frac{2}{y}+\frac{\frac{9}{5}}{y+2}+\frac{\frac{1}{5}}{y-3})$
$=\left[2\ln|y|+\dfrac{9}{5}\ln|y+2|+\dfrac{1}{5}\ln|y-3|\right]_{1}^{2}$
$=2\displaystyle \ln 2+\frac{9}{5}\ln 4+0-(0+\frac{9}{5}\ln 3+\frac{1}{5} \ln 2)$
$=2\displaystyle \ln 2+\frac{18}{5}\ln 2-\frac{1}{5}\ln 2-\frac{9}{5} \ln 3$
$=\displaystyle \frac{27}{5} \ln 2- \displaystyle \frac{9}{5}\ln 3$
$=\displaystyle \frac{27\ln 2-9\ln 3}{5} \approx 1.76549$
