Answer
$$ - \frac{2}{{7\left( {x + 4} \right)}} + \frac{{6x - 3}}{{7\left( {3{x^2} + 1} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{{3x - 2}}{{\left( {x + 4} \right)\left( {3{x^2} + 1} \right)}} \cr
& {\text{The partial fraction decomposition is }} \cr
& \frac{{3x - 2}}{{\left( {x + 4} \right)\left( {3{x^2} + 1} \right)}} = \frac{A}{{x + 4}} + \frac{{Bx + C}}{{3{x^2} + 1}} \cr
& {\text{Multiply each side by }}\left( {x + 4} \right)\left( {3{x^2} + 1} \right) \cr
& 3x - 2 = A\left( {3{x^2} + 1} \right) + \left( {Bx + C} \right)\left( {x + 4} \right)\,\,\,\,\,\,\,\,\left( 1 \right) \cr
& {\text{Expand and combine like terms on the right}} \cr
& 3x - 2 = 3A{x^2} + A + B{x^2} + 4Bx + Cx + 4C \cr
& 3x - 2 = \left( {3A + B} \right){x^2} + \left( {4B + C} \right)x + \left( {A + 4C} \right) \cr
& {\text{Equating the coefficients}} \cr
& A + 4C = - 2 \cr
& 4B + C = 3 \cr
& 3A + B = 0 \cr
& {\text{Solving the system of equations we obtain}} \cr
& A = - \frac{2}{7},\,\,\,B = \frac{6}{7},\,\,\,C = - \frac{3}{7} \cr
& {\text{Use }}A,{\text{ }}B,\,\,{\text{ and }}C{\text{ to find the partial fraction decomposition}} \cr
& \frac{{3x - 2}}{{\left( {x + 4} \right)\left( {3{x^2} + 1} \right)}} = - \frac{2}{{7\left( {x + 4} \right)}} + \frac{{6x - 3}}{{7\left( {3{x^2} + 1} \right)}} \cr} $$
