Answer
$$\int^8_4\frac{ydy}{y^2-2y-3}=\frac{1}{4}\ln9+\frac{1}{2}\ln5=\frac{\ln 15}{2}$$
Work Step by Step
$$A=\int^8_4\frac{ydy}{y^2-2y-3}$$
1) Express the integrand as a sum of partial fractions:
$$\frac{y}{y^2-2y-3}=\frac{y}{(y+1)(y-3)}=\frac{A}{y+1}+\frac{B}{y-3}$$
Clear fractions:
$$y=Ay-3A+By+B$$ $$y=(A+B)y-(3A-B)$$
Equating coefficients of corresponding powers of $y$, we get
- $A+B=1$
- $3A-B=0$
Calculation gives us $A=1/4$ and $B=3/4$.
Therefore, $$\frac{y}{y^2-2y-3}=\frac{1}{4(y+1)}+\frac{3}{4(y-3)}$$
2) Evaluate the integral: $$A=\frac{1}{4}\int^8_4\frac{dy}{y+1}+\frac{3}{4}\int^8_4\frac{dy}{y-3}$$ $$A=\frac{1}{4}\ln|y+1|\Big]^8_4+\frac{3}{4}\ln|y-3|\Big]^8_4$$ $$A=\frac{1}{4}(\ln9-\ln5)+\frac{3}{4}(\ln5-\ln1)$$ $$A=\frac{1}{4}\ln9+\frac{1}{2}\ln5=\frac{\ln 15}{2}$$
