Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {\frac{{{x^3} - 4x}}{{{x^2} + 1}}} dx \cr
& {\text{Identify the symmetry in the function }}\frac{{{x^3} - 4x}}{{{x^2} + 1}},{\text{evaluate }}f\left( { - x} \right) \cr
& f\left( { - x} \right) = \frac{{{{\left( { - x} \right)}^3} - 4\left( { - x} \right)}}{{{{\left( { - x} \right)}^2} + 1}} \cr
& f\left( { - x} \right) = \frac{{ - {x^3} + 4x}}{{{x^2} + 1}} \cr
& f\left( { - x} \right) = \frac{{ - \left( {{x^3} - 4x} \right)}}{{{x^2} + 1}} \cr
& f\left( { - x} \right) = - \frac{{{x^3} - 4x}}{{{x^2} + 1}} \cr
& f\left( { - x} \right) = - f\left( x \right) \cr
& {\text{Therefore }}\frac{{{x^3} - 4x}}{{{x^2} + 1}}{\text{ is an odd function}}{\text{.}} \cr
& {\text{Use the integral property }}\int_{ - a}^a {f\left( x \right)} dx = 0,{\text{ }}f{\text{ is odd}}{\text{.}} \cr
& \int_{ - 2}^2 {\frac{{{x^3} - 4x}}{{{x^2} + 1}}} dx = 0 \cr} $$
