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: 5



Work Step by Step

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