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