Precalculus (6th Edition) Blitzer

The partial fraction decomposition of the rational expression is $\frac{2}{x-2}+\frac{5}{{{\left( x-2 \right)}^{2}}}$
We see that the linear factor $x-2$ occurs multiple times in the denominator of the rational expression, so, for each power of $x-2$ assign an undefined constant as given below: $\frac{2x+1}{{{\left( x-2 \right)}^{2}}}=\frac{A}{x-2}+\frac{B}{{{\left( x-2 \right)}^{2}}}$ Then, multiply both sides of the equation by ${{\left( x-2 \right)}^{2}}$: \begin{align} & {{\left( x-2 \right)}^{2}}\cdot \frac{2x+1}{{{\left( x-2 \right)}^{2}}}={{\left( x-2 \right)}^{2}}\left( \frac{A}{x-2}+\frac{B}{{{\left( x-2 \right)}^{2}}} \right) \\ & {{\left( x-2 \right)}^{2}}\cdot \frac{2x+1}{{{\left( x-2 \right)}^{2}}}={{\left( x-2 \right)}^{2}}\cdot \frac{A}{x-2}+{{\left( x-2 \right)}^{2}}\cdot \frac{B}{{{\left( x-2 \right)}^{2}}} \end{align} And divide the common factors: \begin{align} & {{\left( x-2 \right)}^{2}}\cdot \frac{2x+1}{{{\left( x-2 \right)}^{2}}}=\left( x-2 \right)\left( x-2 \right)\cdot \frac{A}{x-2}+{{\left( x-2 \right)}^{2}}\cdot \frac{B}{{{\left( x-2 \right)}^{2}}} \\ & 2x+1=\left( x-2 \right)A+B \\ & 2x+1=\left( A \right)x+\left( -2A+B \right) \end{align} Then, equate the coefficients of like terms of the equation to write a system of equations as given below: $A=2$ (I) $-2A+B=1$ (II) Put the value of $A$ in equation (II): \begin{align} & -2A+B=1 \\ & -2\left( 2 \right)+B=1 \\ & B=1+4 \\ & B=5 \end{align} Put the value of $A\text{ and }B$ in the initial equation, and define the partial fraction decomposition: \begin{align} & \frac{2x+1}{{{\left( x-2 \right)}^{2}}}=\frac{A}{x-2}+\frac{B}{{{\left( x-2 \right)}^{2}}} \\ & =\frac{2}{x-2}+\frac{5}{{{\left( x-2 \right)}^{2}}} \end{align} Thus, the partial fraction decomposition of the rational expression is $\frac{2}{x-2}+\frac{5}{{{\left( x-2 \right)}^{2}}}$.