assignment submission\nfor this assignment, you submit answers by question parts. the number of…

Explanation:

Step1: Decompose into partial - fractions

Let $\frac{4y^{2}-5y - 12}{y(y + 2)(y - 3)}=\frac{A}{y}+\frac{B}{y + 2}+\frac{C}{y - 3}$. Then $4y^{2}-5y - 12=A(y + 2)(y - 3)+By(y - 3)+Cy(y + 2)$. If $y = 0$, we have $-12=A(2)(-3)$, so $A = 2$. If $y=-2$, we have $4\times4+10 - 12=B(-2)(-5)$, so $B = 3$. If $y = 3$, we have $4\times9-15 - 12=C(3)(5)$, so $C=-1$. So $\frac{4y^{2}-5y - 12}{y(y + 2)(y - 3)}=\frac{2}{y}+\frac{3}{y + 2}-\frac{1}{y - 3}$.

Step2: Integrate term - by - term

$\int_{1}^{2}\frac{4y^{2}-5y - 12}{y(y + 2)(y - 3)}dy=\int_{1}^{2}(\frac{2}{y}+\frac{3}{y + 2}-\frac{1}{y - 3})dy$ $=2\int_{1}^{2}\frac{1}{y}dy+3\int_{1}^{2}\frac{1}{y + 2}dy-\int_{1}^{2}\frac{1}{y - 3}dy$. $=2[\ln|y|]{1}^{2}+3[\ln|y + 2|]{1}^{2}-[\ln|y - 3|]_{1}^{2}$.

Step3: Evaluate the definite integrals

$2(\ln2-\ln1)+3(\ln4-\ln3)-(\ln1-\ln2)$ $=2\ln2+3\ln4 - 3\ln3+\ln2$ $=3\ln2+3\ln4 - 3\ln3$ $=3(\ln2+\ln4-\ln3)$ $=3(\ln(2\times4)-\ln3)$ $=3\ln\frac{8}{3}$.

Answer:

$3\ln\frac{8}{3}$