$\underset{0}{\overset{\infty }{\int }}\frac{x^3}{{(1+x^2)}^{9/2}}dx=$

$$\begin{gathered} {\int }_0^{\infty \\ } \frac{{\mathrm{x}}^3}{{(1+{\mathrm{x}}^2)}^{9/2}} \mathrm{dx} \\ ={\int }_0^{\mathrm{\pi }/2} \frac{{\tan }^3\mathrm{\theta }}{{\left({\sec }^2\mathrm{\theta }\right)}^{9/2}}{\sec }^2\mathrm{\theta } \mathrm{d\theta } \\ \mathrm{put} \mathrm{x}=\mathrm{tan\theta } \Rightarrow \mathrm{dx}={\sec }^2 \mathrm{\theta } \mathrm{d\theta } \\ \mathrm{x}=0 \Rightarrow \theta =0 \\ \mathrm{x}=\infty \Rightarrow \mathrm{\theta }=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ ={\int }_0^{\mathrm{\pi }/2} \frac{{\tan }^3\mathrm{\theta }}{{\sec }^7 \mathrm{\theta }} \mathrm{d\theta } \\ ={\int }_0^{\mathrm{\pi }/2} {\sin }^3\mathrm{\theta }.{\mathrm{co}}^4\mathrm{\theta } \mathrm{d\theta } \\ =\frac{\left(2\right) (3.1)}{7.5.3.1} \end{gathered}$$