Answer
Divergent
Work Step by Step
Since, we have $ \int_1^\infty\dfrac{x}{x^2+1}dx$
Suppose $x^2+1= p \implies dp=2xdx$
or, $\dfrac{1}{2}\int_2^\infty\dfrac{dp}{4}=\dfrac{1}{2} \lim\limits_{a \to \infty} [\ln p]_{2}^{a}$
or, $ =\dfrac{1}{2} \lim\limits_{a \to \infty} (\ln a-\ln 2)$
or, $=\infty$
Hence, the series is Divergent