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.



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


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