Precalculus (6th Edition) Blitzer

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

Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 842: 47


The partial fraction is $\frac{1}{2c\left( x-c \right)}-\frac{1}{2c\left( x+c \right)}$.

Work Step by Step

By the partial fraction of $\frac{1}{{{x}^{2}}-{{c}^{2}}}=\frac{1}{\left( x-c \right)\left( x+c \right)}$ It can be written as, $\frac{1}{\left( x-c \right)\left( x+c \right)}=\frac{A}{x-c}+\frac{B}{x+c}$ …… (1) Multiply both sides by $\left( x-c \right)\left( x+c \right)$, $\begin{align} & 1=A\left( x+c \right)+B\left( x-c \right) \\ & 1=Ax+Ac+Bx-Bc \\ & 1=\left( A+B \right)x+\left( A-B \right)c \\ \end{align}$ And equate the terms, $\begin{align} & A+B=0 \\ & \left( A-B \right)c=1 \end{align}$ Then solving above equation, $A=\frac{1}{2c}\,\,\text{ and }\,B=-\frac{1}{2c}$ Put these values in equation (1), $\begin{align} & \frac{1}{\left( x-c \right)\left( x+c \right)}=\frac{A}{x-c}+\frac{B}{x+c} \\ & =\frac{1}{2c\left( x-c \right)}-\frac{1}{2c\left( x+c \right)} \end{align}$ Thus, the partial fraction decomposition of the rational expression is $\frac{1}{2c\left( x-c \right)}-\frac{1}{2c\left( x+c \right)}$.
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