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Decide if the integral

$$ \int_{-1}^{\infty} \frac{dx}{x^2 + \sqrt[3]{x^4 + 1}}$$

converges.

I decided to write $ \int_{-1}^{\infty} \frac{dx}{x^2 + \sqrt[3]{x^4 + 1}}$ = $ \int_{-1}^{1} \frac{dx}{x^2 + \sqrt[3]{x^4 + 1}}$ + $ \int_{1}^{\infty} \frac{dx}{x^2 + \sqrt[3]{x^4 + 1}}$

It is easy to show that the second integral converges using inequality. But about the first one, how can I argue that it does converges? My attempt was to finding a function bigger than this one that has an easy integral to calculate (if the function is continuous at the interval, I can just take a constant one that equals to 1+ the local maximum!). But I was wondering: Is that really necessary?

Thanks in advance!

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  • 2
    $\begingroup$ The first one is a common Riemann integral of a continuous function ...! $\endgroup$ – DonAntonio Dec 4 '13 at 15:04
  • $\begingroup$ Indeed :P Thanks for the remember! $\endgroup$ – Giiovanna Dec 4 '13 at 15:07
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Hint:

When $x\to \infty$ then $\frac{1}{x^2+\sqrt[3]{1+x^4}}\sim\frac{1}{x^2} $

Hence converge integral!

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$$x^2+(x^4+1)^\frac{1}{3}=0$$ does not have a root for any $x$.

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