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I would like to prove the convergence of the following improper integral:

$$\int_1^2 {\frac{\sqrt{1+x^2}}{\sqrt[3]{16-x^4}}} dx\quad\quad $$

I tried to find antiderivative (with assistance of online calculators) and then check the limits, but was unable to do so. Can anyone help with some direction? Thanks.

Avi Tal
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  • Indefinite integral is a synonym for antiderivative, so there is no sense to talking about convergence of an indefinite integral. Presumably, you are talking about an integral where the function goes to infinity as $x\to2.$ My brain is fried, and I'm blanking on the term for that, but it is no "indefinite." – Thomas Andrews Sep 20 '23 at 10:16
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    Ah, yes, "improper integral" is the term you are seeking. – Thomas Andrews Sep 20 '23 at 10:19
  • @ThomasAndrews Thanks for the correction – Avi Tal Sep 20 '23 at 10:24

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