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