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Integral Help

  1. Aug 21, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫√(1+x2)/x dx


    2. Relevant equations



    3. The attempt at a solution

    Let u = 1+x2
    du = 2xdx
    1/2du=xdx
    x=√(u-1)

    ∫√1+x2/x dx
    =∫√u/√(u-1) du

    or is it 1/2∫√udu as xdx would remove it. This is where I got confused.
     
  2. jcsd
  3. Aug 21, 2012 #2

    SammyS

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    What did you do with the x in the denominator of
    [itex]\displaystyle \int \frac{\sqrt{1+x^2}}{x}\,dx \ ?[/itex]​

    A better substitution would be a trig substitution such as x = tan(θ).
     
  4. Aug 21, 2012 #3
    In my solution attempt, I tried two different ways. In your specific question, I replaced the x in the denominator with √(u-1) , as solving for x in u=1+x^2
     
  5. Aug 22, 2012 #4

    SammyS

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    You're missing an x.

    You have du = 2x dx . Therefore dx = 1/(2x) du.

    This gives you an x2 in the denominator. x2 = u - 1 .
     
  6. Aug 22, 2012 #5
    A trigonometric substitution does the trick here, an hyperbolic substitution might also work.
     
  7. Aug 22, 2012 #6
    Thanks for the help so far. With that little fix, I've now reached the point of

    1/2 ∫√u/(u-1) du

    I don't know where to go from here. I also am having many people tell me that this method will not work? That Trig-Substitution is the only way to solve this problem... Is that true? Am I just wasting my time?
     
  8. Aug 22, 2012 #7

    SammyS

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    There are two ways that I know of to do this integration.

    Use either a trig substitution, such as x = tan(θ) , or use a substitution involving hyperbolic functions, such as x = sinh(u).

    These are handy because, tan2(θ) + 1 = sec2(θ) and sinh2(u) + 1 = cosh2(u).


    Either substitution will require you to do a far amount of follow-up work to finish the integration.
     
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