Answer
$$\frac{{{x^2}}}{2} - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + C{\text{ }}$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{1 + {x^2}}}} dx \cr
& \cr
& {\text{By the long division }}\frac{{{x^3}}}{{1 + {x^2}}} = x - \frac{x}{{{x^2} + 1}}.{\text{Then}} \cr
& \int {\frac{{{x^3}}}{{1 + {x^2}}}} dx = \int {\left( {x - \frac{x}{{{x^2} + 1}}} \right)} dx \cr
& = \int x dx - \int {\frac{x}{{{x^2} + 1}}} dx \cr
& = \int x dx - \frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}} dx \cr
& {\text{Integrating}} \cr
& = \frac{{{x^2}}}{2} - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + C{\text{ }} \cr} $$
