Answer
$$\infty $$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{{{x^3}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ so}} \cr
& \int_0^\infty {\frac{{{x^3}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{{x^3}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx} \cr
& {\text{Integrating }}\int {\frac{{{x^3}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx} \cr
& {\text{Let }}u = {x^2} + 1,{\text{ }}du = 2xdx,{\text{ }}dx = \frac{1}{{2x}}du \cr
& \int {\frac{{{x^3}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx} = \int {\frac{{{x^3}}}{{{u^2}}}\left( {\frac{1}{{2x}}} \right)du} = \int {\frac{{{x^2}}}{{{u^2}}}\left( {\frac{1}{2}} \right)du} \cr
& = \frac{1}{2}\int {\frac{{u - 1}}{{{u^2}}}du} = \frac{1}{2}\int {\left( {\frac{1}{u} - \frac{1}{{{u^2}}}} \right)} du \cr
& = \frac{1}{2}\ln \left| u \right| + \frac{1}{u} + C \cr
& = \frac{1}{2}\ln \left( {{x^2} + 1} \right) + \frac{1}{{{x^2} + 1}} + C \cr
& \cr
& \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{{x^3}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx} = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{2}\ln \left( {{x^2} + 1} \right) + \frac{1}{{{x^2} + 1}}} \right]_0^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{2}\ln \left( {{b^2} + 1} \right) + \frac{1}{{{b^2} + 1}} - \frac{1}{2}\ln \left( {{0^2} + 1} \right) - \frac{1}{{{0^2} + 1}}} \right] \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{2}\ln \left( {{b^2} + 1} \right) + \frac{1}{{{b^2} + 1}} - 1} \right] \cr
& {\text{Evaluate the limit}} \cr
& = \frac{1}{2}\ln \left( {{\infty ^2} + 1} \right) + \frac{1}{{{\infty ^2} + 1}} - 1 \cr
& = \frac{1}{2}\left( \infty \right) + 0 - 1 \cr
& = \infty \cr} $$
