## Precalculus (6th Edition) Blitzer

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