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: 4



Work Step by Step

$\frac{2x+2}{x^{2}-2x+1}=\frac{2x+2}{(x-1)^{2}}$ Decomposing to partial fractions, we get $\frac{2x+2}{(x-1)^{2}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}$ This gives A(x-1)+B= 2x+2. Equating the coefficients of x and the constant term, we get: A=2, -A+B=2 or B=4 Thus we can write $\frac{2x+2}{x^{2}-2x+1}=\frac{2}{x-1}+\frac{4}{(x-1)^{2}}$
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