Answer
$$f\left( x \right) = x{\left( {{x^2} + 1} \right)^2}{\text{ is an odd function}}$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {x{{\left( {{x^2} + 1} \right)}^2}} dx = 0 \cr
& {\text{Let }}f\left( x \right) = x{\left( {{x^2} + 1} \right)^2} \cr
& {\text{Evaluate }}f\left( { - x} \right),{\text{ then}} \cr
& f\left( { - x} \right) = - x{\left( {{{\left( { - x} \right)}^2} + 1} \right)^2} \cr
& f\left( { - x} \right) = - x{\left( {{x^2} + 1} \right)^2} \cr
& f\left( { - x} \right) = - \underbrace {x{{\left( {{x^2} + 1} \right)}^2}}_{f\left( x \right)} \cr
& f\left( { - x} \right) = - f\left( x \right) \cr
& {\text{Therefore }}f\left( x \right) = x{\left( {{x^2} + 1} \right)^2}{\text{ is an odd function}} \cr
& {\text{Using the property }}\int_{ - a}^a {f\left( x \right)} dx = 0,{\text{ }} \cr
& {\text{If }}f\left( x \right){\text{ is an odd function}}{\text{.}} \cr} $$
