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Convergence of integral

  1. Jan 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Find whether the integral is convergent or not, and evaluate if convergent.

    2. Relevant equations
    integral 1/sqrt(x^4+x^2+1) from 1 to infinity

    3. The attempt at a solution
    1/sqrt(x^4+x^2+1)<1/sqrt(x^4)
    1/sqrt(x^4)=1/x^2 which is convergent for 1 to infinity and is 1
    therefore, the original function is convergent
    I can't seem to find the antiderivative of the original function after this.
     
  2. jcsd
  3. Jan 30, 2015 #2
    Try completing the square
     
  4. Jan 30, 2015 #3
    Ok, so I complete the square and got [tex] \int \frac{1}{((x^{2}+\frac{1}{2})^{2}+\frac{3}{4})^{\frac{1}{2}}} [/tex] . Im still not seeing a way find an antiderivative though.
     
  5. Jan 30, 2015 #4

    Mark44

    Staff: Mentor

    Don't forget the dx.
    $$\int \frac{dx}{((x^{2}+\frac{1}{2})^{2}+\frac{3}{4})^{\frac{1}{2}}}$$

    I haven't worked this through, yet, but here's what I would do, FWIW.
    Let u = x2 + 1/2
    So ##x = \sqrt{u - 1/2}##
    Then du = 2xdx
    So ##dx = \frac{du}{2x} = \frac{du}{2\sqrt{u - 1/2}}##

    After making the substitution and getting everything in terms of u and du, I would try for a trig substitution. I think that might work.
     
  6. Jan 30, 2015 #5
    You don't need to integrate the exact expression. If the 3/4 wasn't in there, would the integral be convergent?

    Chet
     
    Last edited: Jan 30, 2015
  7. Jan 30, 2015 #6
    Trig substitution is always my first instinct when I see a fraction with higher powers of x.
     
  8. Jan 31, 2015 #7

    Mark44

    Staff: Mentor

    @chet, I think he already established that in the opening post.
     
  9. Jan 31, 2015 #8
    What I first saw was rewriting the expression as $${\frac{1}{(x^4+x^2+1)^{1/2}}}={\frac{1}{((x^2+({\frac{1}{2}))^2}+(\frac{\sqrt{3}}{2})^2)^{1/2}}}$$.
    Which I'm pretty sure can be solved with a trig sub.
     
  10. Jan 31, 2015 #9
    Ooops. Missed that. Sorry guys.

    Chet
     
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