Answer
The partial fraction is $\frac{b+ac}{2c\left( x-c \right)}+\frac{ac-b}{2c\left( x+c \right)}$.
Work Step by Step
The partial fraction of $\frac{ax+b}{{{x}^{2}}-{{c}^{2}}}=\frac{ax+b}{\left( x-c \right)\left( x+c \right)}$.
It can be written as,
$\frac{ax+b}{\left( x-c \right)\left( x+c \right)}=\frac{A}{x-c}+\frac{B}{x+c}$ …… (1)
And multiply both sides by $\left( x-c \right)\left( x+c \right)$,
$\begin{align}
& ax+b=A\left( x+c \right)+B\left( x-c \right) \\
& ax+b=Ax+Ac+Bx-Bc \\
& ax+b=\left( A+B \right)x+\left( A-B \right)c \\
\end{align}$
And equate the terms,
$\begin{align}
& A+B=a \\
& \left( A-B \right)c=b
\end{align}$
And by solving the above equation,
$A=\frac{b+ac}{2c},B=\frac{ac-b}{2c}$
Put these values in equation (1),
$\frac{ax+b}{\left( x-c \right)\left( x+c \right)}=\frac{b+ac}{2c\left( x-c \right)}+\frac{ac-b}{2c\left( x+c \right)}$
Thus, the partial fraction decomposition is $\frac{b+ac}{2c\left( x-c \right)}+\frac{ac-b}{2c\left( x+c \right)}$.
