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: 14

Answer

$$\int^1_{1/2}\frac{y+4}{y^2+y}dy=3\ln3-2\ln2=\ln \frac{27}{4}$$

Work Step by Step

$$A=\int^1_{1/2}\frac{y+4}{y^2+y}dy$$ 1) Express the integrand as a sum of partial fractions: $$\frac{y+4}{y^2+y}=\frac{y+4}{(y+1)y}=\frac{A}{y+1}+\frac{B}{y}$$ Clear fractions: $$y+4=Ay+By+B$$ $$y+4=(A+B)y+B$$ Equating coefficients of corresponding powers of $y$, we get - $A+B=1$ - $B=4$ That means $A=-3$. Therefore, $$\frac{y+4}{y^2+y}=\frac{4}{y}-\frac{3}{y+1}$$ 2) Evaluate the integral: $$A=4\int^1_{1/2}\frac{dy}{y}-3\int^1_{1/2}\frac{dy}{y+1}$$ $$A=4\ln|y|\Big]^1_{1/2}-3\ln|y+1|\Big]^1_{1/2}$$ $$A=4(\ln1-\ln\frac{1}{2})-3(\ln2-\ln\frac{3}{2})$$ $$A=4(\ln1-\ln1+\ln2)-3(\ln2-\ln3+\ln2)$$ $$A=4\ln2-6\ln2+3\ln3$$ $$A=3\ln3-2\ln2=\ln \frac{27}{4}$$
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