Answer
Equate the expression with $\frac{Ax+B}{a{{x}^{2}}+bx+c}+\frac{Cx+D}{{{\left( a{{x}^{2}}+bx+c \right)}^{2}}}$ and find the values of A and B by comparing the terms
Work Step by Step
In order to find partial fraction decomposition of a rational function with a repeated, prime quadratic factor in the denominator, equate the expression with $\frac{Ax+B}{a{{x}^{2}}+bx+c}+\frac{Cx+D}{{{\left( a{{x}^{2}}+bx+c \right)}^{2}}}$ and find the values of A and B by comparing the terms.
Let us take an example
Consider $\frac{4x}{{{\left( {{x}^{2}}+5 \right)}^{2}}}$
And the partial fraction decomposition of the rational function is:
$\frac{4x}{{{\left( {{x}^{2}}+5 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}+5}+\frac{Cx+D}{{{\left( {{x}^{2}}+5 \right)}^{2}}}$
And multiply both sides by ${{\left( {{x}^{2}}+5 \right)}^{2}}$:
$\begin{align}
& 4x=\left( Ax+B \right)\left( {{x}^{2}}+5 \right)+\left( Cx+D \right) \\
& 4x=A{{x}^{3}}+5Ax+B{{x}^{2}}+5B+Cx+D \\
& 4x=A{{x}^{3}}+B{{x}^{2}}+\left( 5A+C \right)x+D
\end{align}$
And equating the equation:
$\begin{align}
& A=0 \\
& B=0 \\
& 5A+C=4 \\
& D=0
\end{align}$
After that solving the above equation,
$A=0,B=0,C=4,D=0$
Put these values in $\frac{4x}{{{\left( {{x}^{2}}+5 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}+5}+\frac{Cx+D}{{{\left( {{x}^{2}}+5 \right)}^{2}}}$.
Thus, $\frac{4x}{{{\left( {{x}^{2}}+5 \right)}^{2}}}=\frac{4x}{{{\left( {{x}^{2}}+5 \right)}^{2}}}$.
