Answer
Divergent
Work Step by Step
We know that $n^3+n \leq 2n^3, n \geq 0$, this implies that $\frac{1}{n^3+n} \geq \frac{1}{2n^3}$
Then we have
$\sum_{n=1}^{\infty}\frac{\sqrt{n^4+1}}{n^3+n} > \sum_{n=1}^{\infty}\frac{\sqrt{n^4+1}}{2n^3} > \sum_{n=1}^{\infty}\frac{\sqrt{n^4}}{2n^3}=\sum_{n=1}^{\infty}\frac{n^2}{2n^3}=\sum_{n=1}^{\infty}\frac{1}{2n}=\infty$
Thus, by Direct Comparison Test, the series is divergent