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Weird Integral

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data

    An Integral : [tex]

    \int \frac{1}{x^n(1+x^n)^{1/n}} \;\mathrm{d}x}


    2. Relevant equations

    The Standard integrals.

    3. The attempt at a solution

    I'm aware that integrals like this become very easy after a clever substitution...but maybe I'm not that clever :frown: so I can't even start it. If anyone shows me the first step I'll try to take it from there.
    Last edited: Feb 3, 2010
  2. jcsd
  3. Feb 3, 2010 #2


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    Homework Helper

    Fixed your latex:

    \int \frac{1}{x^n(1+x^n)^{1/n}} \mathrm{d}x
    Last edited: Feb 3, 2010
  4. Feb 3, 2010 #3
    You've left the dx at the bottom :wink:
    but how will it become [itex](1+x^n)^{n}[/itex]??? Bringing it to the top will change the sign of the exponent, right?
  5. Feb 3, 2010 #4


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    Homework Helper

    Yes, I got confused. Sorry about that.
  6. Feb 3, 2010 #5
    I tried letting xn = t, but that ended up with [tex]
    \frac{1}{n} \int \frac{1}{\sqrt[n]{t^2+t}}\:\mathrm{d}t
    [/tex], And I don't see how to do it.
    Then I tried letting xn+1 = tn, and got something similarly unsolvable. Can anyone tell me what's the right substitution in this case?
    Last edited: Feb 3, 2010
  7. Feb 4, 2010 #6

    Char. Limit

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    Gold Member

    Here's something:

    [tex]\int x^{-n}(1+x^n)^{-1/n} \mathrm{d}x[/tex]

    And integrate by parts from there. Dunno if it works, though. I tried it on W-A and they used a weird substitution within a substitution.
  8. Feb 4, 2010 #7
    I did that, letting (1+xn)-1/n as first function, and I ended up with :
    \frac{x^{1-n}}{1-n}\,(1+x^n)^{-1/n} + \frac{x}{1-n} + \frac{x^{n+1}}{1-n^2} + C
    Is that correct?
  9. Feb 4, 2010 #8

    Char. Limit

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    Gold Member

    Maybe. I'm too tired to check now. It looks right.
  10. Feb 5, 2010 #9
    Will anyone please confirm if my answer is correct or not? This problem's been bugging me for quite some time.
  11. Feb 5, 2010 #10
  12. Feb 5, 2010 #11
    try trigonometric substitution
  13. Feb 6, 2010 #12
    What trig substitution can you make? PM me if the OP wants to work it out themselves.
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