Answer
$0$
Work Step by Step
Define $f(x)=\frac{x}{x^2+1}$.
$f(-x)=\frac{-x}{(-x)^2+1}=-\frac{x}{x^2+1}=-f(x)$
This shows that $f(x)$ is an odd function.
Then, for any positive number $t$ we have
$\int_{-t}^t\frac{x}{x^2+1}dx=0$
Consequently,
$\int_{-\infty}^{\infty}\frac{x}{x^2+1}dx=\lim\limits_{t \to \infty}\int_{-t}^t\frac{x}{x^2+1}dx=\lim\limits_{t \to \infty}0=0$
