Answer
$$\frac{1}{x} - \frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& \frac{{{x^4} + 1}}{{x{{\left( {{x^2} + 1} \right)}^2}}} \cr
& {\text{The partial fraction decomposition is }} \cr
& \frac{{{x^4} + 1}}{{x{{\left( {{x^2} + 1} \right)}^2}}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& {\text{Multiply each side by }}{x^2}\left( {{x^2} + 5} \right) \cr
& {x^4} + 1 = A{\left( {{x^2} + 1} \right)^2} + x\left( {{x^2} + 1} \right)\left( {Bx + C} \right) + x\left( {Dx + E} \right) \cr
& {\text{Expand and combine like terms on the right of }}\left( 1 \right) \cr
& {x^4} + 1 = A\left( {{x^4} + 2{x^2} + 1} \right) + \left( {{x^3} + x} \right)\left( {Bx + C} \right) + \left( {D{x^2} + Ex} \right) \cr
& {x^4} + 1 = A{x^4} + 2A{x^2} + A + B{x^4} + C{x^3} + B{x^2} + Cx + D{x^2} + Ex \cr
& {x^4} + 1 = A{x^4} + B{x^4} + C{x^3} + 2A{x^2} + B{x^2} + D{x^2} + Cx + Ex + A \cr
& {x^4} + 1 = \left( {A + B} \right){x^4} + C{x^3} + \left( {2A + B + D} \right){x^2} + \left( {C + E} \right)x + A \cr
& {\text{Equating the coefficients}} \cr
& A + B = 1 \cr
& C = 0 \cr
& 2A + B + D = 0 \cr
& C + E = 0 \cr
& A = 1 \cr
& {\text{Solving the system of equations we obtain}} \cr
& A = 1,\,\,\,B = 0,\,\,\,C = 0,\,\,\,\,D = - 2,\,\,\,E = 0 \cr
& {\text{Use }}A,{\text{ }}B,\,C,D\,\,{\text{ and }}E{\text{ to find the partial fraction decomposition}} \cr
& \frac{{{x^4} + 1}}{{x{{\left( {{x^2} + 1} \right)}^2}}} = \frac{1}{x} - \frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr} $$
