University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.4 - Integration of Rational Functions by Partial Fractions - Exercises - Page 445: 13


$$\int^8_4\frac{ydy}{y^2-2y-3}=\frac{1}{4}\ln9+\frac{1}{2}\ln5=\frac{\ln 15}{2}$$

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: $$A=\frac{1}{4}\int^8_4\frac{dy}{y+1}+\frac{3}{4}\int^8_4\frac{dy}{y-3}$$ $$A=\frac{1}{4}\ln|y+1|\Big]^8_4+\frac{3}{4}\ln|y-3|\Big]^8_4$$ $$A=\frac{1}{4}(\ln9-\ln5)+\frac{3}{4}(\ln5-\ln1)$$ $$A=\frac{1}{4}\ln9+\frac{1}{2}\ln5=\frac{\ln 15}{2}$$
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