Answer
$$\pi$$
Work Step by Step
The volume is given by
\begin{align*}
V&=2 \pi \int_{0}^{\infty}xf(x)dx\\
&=2 \pi \int_{0}^{\infty} \frac{x}{\left(x^{2}+1\right)^{2}} d x\\
&= 2 \pi\lim _{R \rightarrow \infty} \int_{0}^{R} \frac{x d x}{\left(x^{2}+1\right)^{2}}\\
&=2 \pi \lim _{R \rightarrow \infty} \frac{-1}{2(x^2+1)} \bigg|_{0}^{R}\\
&=2 \pi \lim _{R \rightarrow \infty} \frac{1}{2}\left(1-\frac{1}{R^{2}+1}\right)\\
&=\pi
\end{align*}
