Answer
$\dfrac{1}{(x-c)(x+c)}=\dfrac{1}{2c(x-c)}-\dfrac{1}{2c(x+c)}$
Work Step by Step
We are given the fraction:
$\dfrac{1}{x^2-c^2}$
Factor the denominator:
$\dfrac{1}{x^2-c^2}=\dfrac{1}{(x-c)(x+c)}$
Write the partial fraction decomposition:
$\dfrac{1}{(x-c)(x+c)}=\dfrac{A}{x-c}+\dfrac{B}{x+c}$
Multiply all terms by the least common denominator $(x-c)(x+c)$:
$(x-c)(x+c)\cdot\dfrac{1}{(x-c)(x+c)}=(x-c)(x+c)\cdot\dfrac{A}{x-c}+(x-c)(x+c)\cdot\dfrac{B}{x+c}$
$1=A(x+c)+B(x-c)$
$1=Ax+Ac+Bx-Bc$
$1=(A+B)x+(Ac-Bc)$
Identify the coefficients:
$\begin{cases}
A+B=0\\
Ac-Bc=1
\end{cases}$
$\begin{cases}
-cA-cB=-c(0)\\
Ac-Bc=1
\end{cases}$
$-Ac-Bc+Ac-Bc=0+1$
$-2Bc=1$
$B=-\dfrac{1}{2c}$
$A+B=0$
$A-\dfrac{1}{2c}=$
$A=\dfrac{1}{2c}$
The partial fraction decomposition is:
$\dfrac{1}{(x-c)(x+c)}=\dfrac{\dfrac{1}{2c}}{x-c}+\dfrac{-\dfrac{1}{2c}}{x+c}$
$\dfrac{1}{(x-c)(x+c)}=\dfrac{1}{2c(x-c)}-\dfrac{1}{2c(x+c)}$
