Answer
$$\eqalign{
& {\text{Use }}A,{\text{ }}B,\,\,{\text{ and }}C{\text{ to find the partial fraction decomposition}} \cr
& \frac{1}{{x\left( {2x + 1} \right)\left( {3{x^2} + 4} \right)}} = \cr} $$
Work Step by Step
$$\eqalign{
& \frac{1}{{x\left( {2x + 1} \right)\left( {3{x^2} + 4} \right)}} \cr
& {\text{The partial fraction decomposition is }} \cr
& \frac{1}{{x\left( {2x + 1} \right)\left( {3{x^2} + 4} \right)}} = \frac{A}{x} + \frac{B}{{2x + 1}} + \frac{{Cx + D}}{{3{x^2} + 4}} \cr
& {\text{Multiply each side by }}\left( {x + 4} \right)\left( {3{x^2} + 1} \right) \cr
& 1 = A\left( {2x + 1} \right)\left( {3{x^2} + 4} \right) + Bx\left( {3{x^2} + 4} \right) + x\left( {Cx + D} \right)\left( {2x + 1} \right)\,\,\,\,\,\,\,\,\left( 1 \right) \cr
& {\text{Expand and combine like terms on the right}} \cr
& 1 = 6A{x^3} + 3A{x^2} + 8Ax + 4A + 3B{x^3} + 4Bx + 2C{x^3} + C{x^2} \cr
& \,\,\,\,\,\,\,\, + 2D{x^2} + Dx \cr
& 1 = \left( {6A + 3B + 2C} \right){x^3} + \left( {3A + C\, + 2D} \right){x^2} + \left( {8A + 4B + 4D} \right)x + 4A \cr
& {\text{Equating the coefficients}} \cr
& 6A + 3B + 2C = 0 \cr
& 3A + C\, + 2D = 0 \cr
& 8A + 4B + 4D = 0 \cr
& 4A = 1 \cr
& {\text{Solving the system of equations we obtain}} \cr
& A = \frac{1}{4},\,\,\,B = - \frac{3}{{28}},\,\,\,C = \frac{1}{{28}},\,\,\,D = - \frac{{11}}{{28}} \cr
& {\text{Use }}A,{\text{ }}B,\,\,{\text{ and }}C{\text{ to find the partial fraction decomposition}} \cr
& \frac{1}{{x\left( {2x + 1} \right)\left( {3{x^2} + 4} \right)}} = \frac{1}{{4x}} - \frac{3}{{28\left( {2x + 1} \right)}} + \frac{{x - 11}}{{28\left( {3{x^2} + 4} \right)}} \cr} $$
