Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Mid-Chapter Check Point - Page 853: 16



Work Step by Step

Here, we have $\dfrac{x^3}{(x^2+4)^2}=\dfrac{Ax+B}{x^2+4}+\dfrac{Cx+D}{(x^2+4)^2}$ $x^3=Ax^3+Bx^2+(4A+C)x+4B+D$ Equating the coefficients, we get $A=1$ and $B=0$ and $4A+C=0$ and $4B+D=0$ This gives: $A=1; B=0; C= -4; D=0$ Hence, $\dfrac{x^3}{(x^2+4)^2}=\dfrac{x}{x^2+4}-\dfrac{4x}{(x^2+4)^2}$
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