Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.5 Exercises - Page 549: 1

Answer

$\frac{4}{x^2-8x}$=$\frac{A}{x}$+$\frac{B}{x-8}$

Work Step by Step

Put a constant in the numerator on top of each fraction and simply factor x out leaving you with two linear fractions. $\frac{4}{x^2-8x}$=$\frac{A}{x}$+$\frac{B}{x-8}$
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