Answer
$$ - \frac{1}{{4\left( {x + 1} \right)}} + \frac{{x + 7}}{{4\left( {{x^2} + 2} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{{2x + 1}}{{\left( {x + 1} \right)\left( {{x^2} + 2} \right)}} \cr
& {\text{The partial fraction decomposition is }} \cr
& \frac{{2x + 1}}{{\left( {x + 1} \right)\left( {{x^2} + 2} \right)}} = \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} + 2}} \cr
& {\text{Multiply each side by }}\left( {x + 4} \right)\left( {3{x^2} + 1} \right) \cr
& 2x + 1 = A\left( {{x^2} + 2} \right) + \left( {Bx + C} \right)\left( {x + 1} \right)\,\,\,\,\,\,\,\,\left( 1 \right) \cr
& {\text{Expand and combine like terms on the right}} \cr
& 2x + 1 = A{x^2} + 3A + B{x^2} + Bx + Cx + C \cr
& 2x + 1 = \left( {A{x^2} + B{x^2}} \right) + \left( {Bx + Cx} \right) + 3A + C \cr
& {\text{Equating the coefficients}} \cr
& 3A + C = 1 \cr
& A + B = 0 \cr
& B + C = 2 \cr
& {\text{Solving the system of equations we obtain}} \cr
& A = - \frac{1}{4},\,\,\,B = \frac{1}{4},\,\,\,C = \frac{7}{4} \cr
& {\text{Use }}A,{\text{ }}B,\,\,{\text{ and }}C{\text{ to find the partial fraction decomposition}} \cr
& \frac{{2x + 1}}{{\left( {x + 1} \right)\left( {{x^2} + 2} \right)}} = - \frac{1}{{4\left( {x + 1} \right)}} + \frac{{x + 7}}{{4\left( {{x^2} + 2} \right)}} \cr} $$
