Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Review - Concept Check - Page 537: 2

Answer

The statement is True. $$ \frac{x^{2}+4}{x\left(x^{2}-4\right) } $$ can be put in the form: $$ \frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2} $$ In fact, $A = −1, B = 1~and ~C = 1$

Work Step by Step

Resolve into partial fractions: $$ \frac{x^{2}+4}{x\left(x^{2}-4\right) } $$ We’ll first factor the denominator and then get the form of the partial fraction decomposition. $$ \begin{aligned} \frac{x^{2}+4}{x\left(x^{2}-4\right) }& =\frac{x^2+4}{x\left(x+2\right)\left(x-2\right)}\\ &=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2} \\ &=\frac{A\left(x+2\right)\left(x-2\right)+Bx\left(x-2\right)+Cx\left(x+2\right) }{x\left(x+2\right)\left(x-2\right)} \end{aligned} $$ multiplying both sides by the denominator, we get: $$ x^2+4=A\left(x+2\right)\left(x-2\right)+Bx\left(x-2\right)+Cx\left(x+2\right) $$ Substituting $x = 0 $ into this equation gives $A = −1$, substituting $x = -2 $ gives $ B = 1$, and substituting $x =2 $ gives $ C = 1$ so that $$ \frac{x^{2}+4}{x\left(x^{2}-4\right) }=\frac{-1}{x}+\frac{1}{x+2}+\frac{1}{x-2} $$ Thus, the statement is True. In fact, $A = −1, B = 1, C = 1$
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