Answer
$$\frac{1}{x} - \frac{x}{{{x^2} + 1}}$$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{{{x^3}\left( {{x^2} + 1} \right)}} \cr
& {\text{First simplify}} \cr
& \frac{1}{{x\left( {{x^2} + 1} \right)}} \cr
& {\text{Decompose into partial fractions}} \cr
& \frac{1}{{x\left( {{x^2} + 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} \cr
& 1 = A\left( {{x^2} + 1} \right) + x\left( {Bx + C} \right) \cr
& 1 = A{x^2} + A + B{x^2} + Cx \cr
& {\text{Collect like terms}} \cr
& 1 = \left( {A + B} \right){x^2} + Cx + A \cr
& {\text{Equating coefficients}} \cr
& A + B = 0,{\text{ }}C = 0,{\text{ }}A = 1 \cr
& {\text{Solving the system of equations we obtain}} \cr
& A = 1,{\text{ }}B = - 1,{\text{ }}C = 0,{\text{ then}} \cr
& \underbrace {\frac{1}{{x\left( {{x^2} + 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}}}_ \Downarrow \cr
& \frac{1}{{x\left( {{x^2} + 1} \right)}} = \frac{1}{x} - \frac{x}{{{x^2} + 1}} \cr} $$
