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1. - / 2.08 points

evaluate the integral.
\\(\int_{1}^{2}\frac{4y^{2}-5y - 12}{y(y + 2)(y - 3)}dy\\)

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:

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