Answer
\[ = \frac{A}{x} + \frac{B}{{\,\left( {x - 3} \right)}} + \frac{C}{{\,{{\left( {x - 3} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
\frac{2}{{x\,\left( {{x^2} - 6x + 9} \right)}} \hfill \\
\hfill \\
{\text{factor}}\,\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\
\hfill \\
\frac{2}{{x\,\left( {{x^2} + 6x + 9} \right)}} = \frac{2}{{x\,\left( {{x^2} - 2\,\left( 3 \right)\,\left( x \right) + {3^2}} \right)}} = \frac{2}{{x\,{{\left( {x - 3} \right)}^2}}} \hfill \\
\hfill \\
Therefore, \hfill \\
\hfill \\
we\,can\,\,use\,partial\,fraction\,of\,the\,form \hfill \\
\hfill \\
\frac{2}{{x\,{{\left( {x - 3} \right)}^2}}} = \frac{A}{x} + \frac{B}{{\,\left( {x - 3} \right)}} + \frac{C}{{\,{{\left( {x - 3} \right)}^2}}} \hfill \\
\hfill \\
\end{gathered} \]