## Precalculus (6th Edition) Blitzer

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