Answer
$$3{x^2} + \frac{3}{x} - \frac{{3x}}{{{x^2} + 1}}$$
Work Step by Step
$$\eqalign{
& \frac{{3{x^6} + 3{x^4} + 3x}}{{{x^4} + {x^2}}} \cr
& {\text{By using the long division}} \cr
& \frac{{3{x^6} + 3{x^4} + 3x}}{{{x^4} + {x^2}}} = 3{x^2} + \frac{{3x}}{{{x^4} + {x^2}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3{x^2} + \frac{3}{{{x^3} + x}} \cr
& {\text{Decompose }}\frac{3}{{{x^3} + x}}{\text{ into partial fractions}} \cr
& \frac{3}{{{x^3} + x}} = \frac{3}{{x\left( {{x^2} + 1} \right)}} \cr
& \frac{3}{{x\left( {{x^2} + 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} \cr
& \cr
& {\text{Multiply each side by }}x\left( {{x^2} + 1} \right) \cr
& 3 = A\left( {{x^2} + 1} \right) + x\left( {Bx + C} \right) \cr
& 3 = A{x^2} + A + B{x^2} + Cx \cr
& {\text{Equating the coefficients}} \cr
& A + B = 0 \cr
& C = 0 \cr
& A = 3 \cr
& {\text{Then,}} \cr
& B = - 3 \cr
& {\text{Use }}A,{\text{ }}B,\,\,\,{\text{ and }}C{\text{ to find the partial fraction decomposition}} \cr
& \frac{3}{{x\left( {{x^2} + 1} \right)}} = \frac{3}{x} - \frac{{3x}}{{{x^2} + 1}} \cr
& {\text{Therefore,}} \cr
& 3{x^2} + \frac{3}{{{x^3} + x}} = 3{x^2} + \frac{3}{x} - \frac{{3x}}{{{x^2} + 1}} \cr} $$
