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Integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx

  1. Jul 16, 2015 #1
    • Member warned about posting with no template and no effort shown
    Hi,
    I'm trying to solve ∫(x2-1)/(1+x2)*1/(1+x4)(1/2)dx


    I'm apparently meant to get some non-complex result, the question suggests to use the substitution u2 = x2 + 1/x2
    But I haven't gotten anywhere with this.

    Any methods or suggestions (or the solution) would be much appreciated!
    Thanks
     

    Attached Files:

  2. jcsd
  3. Jul 16, 2015 #2

    RUber

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    If you take the hint, ##u^2 = x^2 + \frac{1}{x^2} ## then ## du = \frac{x - \frac {1}{x^3}}{\sqrt{x^2 + \frac{1}{x^2}}}dx= \frac{x^2 - \frac {1}{x^2}}{\sqrt{x^4 +1}}dx##
    So try to substitute in the du and try simplifying from there.
     
  4. Jul 16, 2015 #3

    Ray Vickson

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    This question was posed here about 2-3 months ago, and was discussed thoroughly (and solved) then. If you look for it you will find it.
     
  5. Jul 17, 2015 #4
    Here's my two attempts at a solution (photo attached - second photo is a little messy) - I tried a trig substitution and the u substitution and a following simplification.

    I also looked through the past 5 months of "integration" related questions and couldn't find this particular question so if you had a direct link to it that would be much appreciated.

    Thanks
     

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  6. Jul 17, 2015 #5
    I THINK I GOT IT. If someone could check the photo for errors that'd be great.
     

    Attached Files:

  7. Jul 17, 2015 #6

    RUber

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    Wasn't x^2+1/x^2= u^2?
    I think you may need an inverse tangent function in that case.
     
  8. Jul 17, 2015 #7
    Right you are! I'll fix that up.

    Should be:
    (1/sqrt2)*arctan(sqrt((x^2 + 1/x^2)/2)) yeah?
     
    Last edited: Jul 17, 2015
  9. Jul 17, 2015 #8

    RUber

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    That is what WolframAlpha.com says.
     
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