Answer
The partial fraction decomposition of the rational expression is $\frac{2}{x-2}+\frac{5}{{{\left( x-2 \right)}^{2}}}$
Work Step by Step
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 $\frac{2}{x-2}+\frac{5}{{{\left( x-2 \right)}^{2}}}$.
