Answer
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
$\frac{1}{2c\left( x-c \right)}-\frac{1}{2c\left( x+c \right)}$.
