## Precalculus (6th Edition) Blitzer

The partial fraction is $\frac{a}{x-c}+\frac{ac+b}{{{\left( x-c \right)}^{2}}}$.
We know that the partial fraction can be written as given below, \begin{align} & \frac{ax+b}{{{\left( x-c \right)}^{2}}}=\frac{A}{x-c}+\frac{B}{{{\left( x-c \right)}^{2}}} \\ & =\frac{A\left( x-c \right)+B}{{{\left( x-c \right)}^{2}}} \end{align} And, $ax+b=A\left( x-c \right)+B$ …… (1) Substitute the value of $x=c$ in equation (1), $ac+b=B$ Again, put the value of $x=0$ in equation (1), \begin{align} & b=-Ac+B \\ & b=-Ac+ac+b \\ & Ac=ac \\ & A=a \end{align} Therefore, $\frac{ax+b}{{{\left( x-c \right)}^{2}}}=\frac{a}{x-c}+\frac{ac+b}{{{\left( x-c \right)}^{2}}}$ Thus, the partial fraction decomposition form is $\frac{ax+b}{{{\left( x-c \right)}^{2}}}=\frac{a}{x-c}+\frac{ac+b}{{{\left( x-c \right)}^{2}}}$.