Answer
$\dfrac{ax+b}{(x-c)(x+c)}=\dfrac{ac+b}{2c(x-c)}+\dfrac{ac-b}{2c(x+c)}$
Work Step by Step
We are given the fraction:
$\dfrac{ax+b}{x^2-c^2}$
Factor the denominator:
$\dfrac{ax+b}{x^2-c^2}=\dfrac{ax+b}{(x-c)(x+c)}$
Write the partial fraction decomposition:
$\dfrac{ax+b}{(x-c)(x+c)}=\dfrac{A}{x-c}+\dfrac{B}{x+c}$
Multiply all terms by the least common denominator $(x-c)(x+c)$:
$(x-c)(x+c)\cdot\dfrac{ax+b}{(x-c)(x+c)}=(x-c)(x+c)\cdot\dfrac{A}{x-c}+(x-c)(x+c)\cdot\dfrac{B}{x+c}$
$ax+b=A(x+c)+B(x-c)$
$ax+b=Ax+Ac+Bx-Bc$
$ax+b=(A+B)x+(Ac-Bc)$
Identify the coefficients:
$\begin{cases}
A+B=a\\
Ac-Bc=b
\end{cases}$
$\begin{cases}
-cA-cB=-ca\\
Ac-Bc=b
\end{cases}$
$-Ac-Bc+Ac-Bc=-ca+b$
$-2Bc=-ca+b$
$B=\dfrac{ac-b}{2c}$
$A+B=a$
$A+\dfrac{ac-b}{2c}=a$
$A=a-\dfrac{ac-b}{2c}$
$A=\dfrac{ac+b}{2c}$
The partial fraction decomposition is:
$\dfrac{ax+b}{(x-c)(x+c)}=\dfrac{\dfrac{ac-b}{2c}}{x-c}+\dfrac{\dfrac{ac+b}{2c}}{x+c}$
$\dfrac{ax+b}{(x-c)(x+c)}=\dfrac{ac+b}{2c(x-c)}+\dfrac{ac-b}{2c(x+c)}$
