## Precalculus (6th Edition) Blitzer

The expression is equated to $\frac{A}{ax+b}+\frac{B}{{{\left( ax+b \right)}^{2}}}$ and then the values of A and B are calculated by comparing terms.
In order to find the partial fraction decomposition of a rational function with a repeated linear factors in the denominator, the expression is equated to $\frac{A}{ax+b}+\frac{B}{{{\left( ax+b \right)}^{2}}}$ and then the values of A and B are calculated by comparing terms. Let us take an example Consider $\frac{2x}{{{\left( x+2 \right)}^{2}}}$ And the partial fraction decomposition of the rational function is: $\frac{2}{{{\left( x+2 \right)}^{2}}}=\frac{A}{x+2}+\frac{B}{{{\left( x+2 \right)}^{2}}}$ And multiply both sides of the equation by ${{\left( x+2 \right)}^{2}}$: \begin{align} & 2x=A\left( x+2 \right)+B \\ & 2x=Ax+2A+B \\ & 2x=Ax+2A+B \end{align} And equate the coefficients of the above equation: $A=2$ (I) $2A+B=0$ (II) Solving the above equation, And put the value of $A=2$ in equation (II): \begin{align} & 2\left( 2 \right)+B=0 \\ & 4+B=0 \\ & B=-4 \end{align} This implies, $A=2,B=-4$. Put these values in $\frac{2}{{{\left( x+2 \right)}^{2}}}=\frac{A}{x+2}+\frac{B}{{{\left( x+2 \right)}^{2}}}$. Thus, $\frac{2}{{{\left( x+2 \right)}^{2}}}=\frac{2}{x+2}-\frac{4}{{{\left( x+2 \right)}^{2}}}$.