Answer
Divergent
Work Step by Step
$\int_1^\infty f(x)dx=\int_1^\infty \frac{x}{x^{2}+1}dx$
Let $u=x^{2}+1$, so $du=2xdx$
$=\int_{u=2}^\infty \frac{du}{u}$
$=\frac{1}{2}\int_{1}^\infty \frac{1}{u}du$
$=\frac{1}{2}lnu|_{1}^\infty$
$=\frac{1}{2}ln(x^{2}+1)|_{1}^\infty$
$=\infty $
Hence, the given series is divergent.