1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.