Answer
$$ - \frac{2}{{9x}} + \frac{2}{{3{x^2}}} + \frac{2}{{9\left( {x + 3} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{2}{{{x^2}\left( {x + 3} \right)}} \cr
& {\text{The partial fraction decomposition is }} \cr
& \frac{2}{{{x^2}\left( {x + 3} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 3}} \cr
& {\text{Multiply each side by }}\left( {x + 1} \right){\left( {x + 2} \right)^2} \cr
& 2 = Ax\left( {x + 3} \right) + B\left( {x + 3} \right) + C{x^2}\,\,\,\,\,\,\,\,\,\,\left( 1 \right) \cr
& \cr
& {\text{Let }}x = 0{\text{ into the equation }}\left( 1 \right) \cr
& 2 = A\left( 0 \right) + B\left( 3 \right) + C\left( 0 \right)\,\, \cr
& B = \frac{2}{3} \cr
& \cr
& {\text{Let }}x = - 3{\text{ into the equation }}\left( 1 \right) \cr
& 2 = A\left( 0 \right) + B\left( 0 \right) + C{\left( { - 3} \right)^2} \cr
& 2 = C{\left( { - 3} \right)^2} \cr
& C = \frac{2}{9} \cr
& \cr
& {\text{Let }}x = 1,\,\,\,C = \frac{2}{9}{\text{ and }}A = - 2{\text{ into the equation }}\left( 1 \right) \cr
& 2 = A\left( 1 \right)\left( {1 + 3} \right) + \frac{2}{3}\left( {1 + 3} \right) + \frac{2}{9}{\left( 1 \right)^2} \cr
& 2 = A\left( 4 \right) + \frac{8}{3} + \frac{2}{9} \cr
& A = - \frac{2}{9} \cr
& \cr
& {\text{Use }}A,{\text{ }}B{\text{ and }}C{\text{ to find the partial fraction decomposition}} \cr
& \frac{2}{{{x^2}\left( {x + 3} \right)}} = - \frac{2}{{9x}} + \frac{2}{{3{x^2}}} + \frac{2}{{9\left( {x + 3} \right)}} \cr} $$
