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: 4

Answer

$\frac{2x-1}{x(x^2+1)^2}= \frac{A}{x} + \frac{Bx+C}{(x^2+1)}+ \frac{Dx+E}{(x^2+1)^2}$

Work Step by Step

Rewrite the fraction using partial fractions Because the denominator is already in simplest terms, begin the decomposition. Note that for quadratic terms an $Ax+B$ (or other constants for A and B) must be used $\frac{2x-1}{x(x^2+1)^2}= \frac{A}{x} + \frac{Bx+C}{(x^2+1)}+ \frac{Dx+E}{(x^2+1)^2}$
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