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Improper integral, divergence/convergence

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate if the integral diverges or converges using the comparison theorem.

    [tex]\int^ \infty_2 \frac{dx}{\sqrt{x^3+1}}[/tex]

    Having trouble with this question, the exercises I have managed I generally guessed if it was convergent/divergent, and then found a smaller of bigger function at the interval and went by that. Here I assumed it was convergent and used [tex]\frac{1}{\sqrt{x^2+1}}[/tex], though that was divergent which says absolutely nothing. Ran into the same issue with other attempts.

    Is my reasoning the issue or have I just not found the correct manipulation? Not really approaching with anything more involved than "Find something bigger and hope it converges".
    Any help would be appreciated, thanks.
     
  2. jcsd
  3. Nov 10, 2012 #2

    HallsofIvy

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    It seems to me that a simpler comparison is [itex]x^3< x^3+ 1[/itex] so that [itex]\frac{1}{\sqrt{x^3+ 1}}< \frac{1}{\sqrt{x^3}}= x^{-3/2}[/itex].
     
    Last edited: Nov 10, 2012
  4. Nov 10, 2012 #3
    I feel slightly moronic now, was too focused on manipulating the x term for whatever reason.

    Thanks, glad I didn't get it all backwards at least.
     
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