Answer
Divergent
Work Step by Step
Plug $x^2+1= a$ and $da=(2x) dx$
Now , $\int_1^\infty\dfrac{x}{x^2+1}dx=(\dfrac{1}{2})\int_2^\infty\dfrac{da}{4}=(\dfrac{1}{2}) [\lim\limits_{k \to \infty} [\ln a]_{2}^{k}$
and $ (\dfrac{1}{2}) \lim\limits_{k \to \infty} [\ln (k)-\ln (2)]=\infty$
Hence, the series is Divergent