Question

express the integrand as a sum of partial fractions and evaluate the integral.
int_{5}^{9}\frac{y}{y^{2}-y - 2}dy
express the integrand as a sum of partial fractions.
\frac{y}{y^{2}-y - 2}=square
(simplify your answer. use integers or fractions for any numbers in the expression.)
evaluate the integral.
int_{5}^{9}\frac{y}{y^{2}-y - 2}dy=square
(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Factor the denominator

Factor $y^{2}-y - 2=(y - 2)(y+1)$.

Step2: Set up partial - fraction decomposition

Let $\frac{y}{y^{2}-y - 2}=\frac{A}{y - 2}+\frac{B}{y + 1}$. Then $y=A(y + 1)+B(y - 2)$.

Step3: Solve for A and B

If $y = 2$, then $2=A(2 + 1)+B(2 - 2)$, so $A=\frac{2}{3}$.
If $y=-1$, then $-1=A(-1 + 1)+B(-1 - 2)$, so $B=\frac{1}{3}$.
So $\frac{y}{y^{2}-y - 2}=\frac{2/3}{y - 2}+\frac{1/3}{y + 1}$.

Step4: Evaluate the integral

$\int_{5}^{9}\frac{y}{y^{2}-y - 2}dy=\int_{5}^{9}(\frac{2/3}{y - 2}+\frac{1/3}{y + 1})dy$
$=\frac{2}{3}\int_{5}^{9}\frac{1}{y - 2}dy+\frac{1}{3}\int_{5}^{9}\frac{1}{y + 1}dy$
$=\frac{2}{3}[\ln|y - 2|]_{5}^{9}+\frac{1}{3}[\ln|y + 1|]_{5}^{9}$
$=\frac{2}{3}(\ln(9 - 2)-\ln(5 - 2))+\frac{1}{3}(\ln(9 + 1)-\ln(5 + 1))$
$=\frac{2}{3}(\ln7-\ln3)+\frac{1}{3}(\ln10-\ln6)$
$=\frac{2}{3}\ln\frac{7}{3}+\frac{1}{3}\ln\frac{10}{6}$
$=\frac{2}{3}\ln\frac{7}{3}+\frac{1}{3}\ln\frac{5}{3}$
$=\ln((\frac{7}{3})^{\frac{2}{3}})+\ln((\frac{5}{3})^{\frac{1}{3}})$
$=\ln((\frac{7}{3})^{\frac{2}{3}}(\frac{5}{3})^{\frac{1}{3}})$
$=\ln(\frac{7^{\frac{2}{3}}\times5^{\frac{1}{3}}}{3})$

Answer:

$\frac{y}{y^{2}-y - 2}=\frac{2/3}{y - 2}+\frac{1/3}{y + 1}$; $\int_{5}^{9}\frac{y}{y^{2}-y - 2}dy=\ln(\frac{7^{\frac{2}{3}}\times5^{\frac{1}{3}}}{3})$