# Improper Integral - Divergence with Comparison Test

Show that the following integral diverges $$\int_{0}^{\infty} \frac{x}{\sqrt{x^3 + 1}} \ dx.$$

This looked like an easy question, but I spent a lot of time on it. I am not sure which test function to use for comparison. Maybe I misunderstood something.

$\frac{1}{\sqrt{x}}$ is bigger than $f$ for positive $x$, so its divergence does not imply the divergence of $f$. I tried $\frac{x}{\sqrt{x^3 + x}}$ but then I have same problem proving its divergence.

Hints?

Hint. Note that for $x\geq 1$, $$\frac{x}{\sqrt{x^3 + 1}}=\frac{1}{\sqrt{x+\frac{1}{x^2}}}\geq \frac{1}{\sqrt{x+1}}.$$