Answer
Divergent
Work Step by Step
\[\begin{align}
\text{Let }& I=\int_{-\infty }^{\infty }{\frac{x}{{{x}^{2}}+1}}dx \\
& \text{Therefore} \\
& \int_{-\infty }^{\infty }{\frac{x}{{{x}^{2}}+1}}dx=2\int_{0}^{\infty }{\frac{x}{{{x}^{2}}+1}}dx \\
& =\int_{0}^{\infty }{\frac{2x}{{{x}^{2}}+1}}dx \\
& \text{Integrating} \\
& I=\left[ \ln \left( {{x}^{2}}+1 \right) \right]_{0}^{\infty } \\
& =\left( \infty \right)-0 \\
& =\infty \\
& \text{The improper integral diverges} \\
& \text{Divergent} \\
\end{align}\]
