#### Answer

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.

#### Work Step by Step

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, $\frac{2}{{{\left( x+2 \right)}^{2}}}=\frac{2}{x+2}-\frac{4}{{{\left( x+2 \right)}^{2}}}$.