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: 6



Work Step by Step

$$\frac{z}{z^3-z^2-6z}=\frac{1}{z^2-z-6}=\frac{1}{(z+2)(z-3)}$$ The partial fraction decomposition then has the following form: $$\frac{z}{z^3-z^2-6z}=\frac{1}{(z+2)(z-3)}=\frac{A}{z+2}+\frac{B}{z-3}$$ To find $A, B$, we clear fractions and get $$1=A(z-3)+B(z+2)$$ $$1=Az-3A+Bz+2B$$ $$1=(A+B)z+(2B-3A)$$ Equating coefficients of corresponding powers of $z$, we get - $A+B=0$ - $2B-3A=1$ So $A=-1/5$ and $B=1/5$ Therefore, $$\frac{z}{z^3-z^2-6z}=-\frac{1}{5(z+2)}+\frac{1}{5(z-3)}$$
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