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Integral of sqrt((x)/(x-1))

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]\int[/itex]([itex]\sqrt{x}/\sqrt{x-1}[/itex] )dx.


    2. Relevant equations
    -


    3. The attempt at a solution

    It should be doable with substitution or/and with partial intergral. I just don't figure out what to substitute. I have tried with u = √(x-1), u = √(x), and with partial integral formula:

    ∫u*v´ = u*v - ∫v * u´

    Any tips?

    Thanks for any help
    -Siune
     
  2. jcsd
  3. Jan 28, 2012 #2

    SammyS

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    Hello Siune. Welcome to PF .

    Try the substitution, u = x-1 .
     
  4. Jan 28, 2012 #3

    HallsofIvy

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    I think much simpler is to let [itex]u= \sqrt{x}[/itex]. Now what are dx and x- 1 in terms of u and du?
     
  5. Jan 28, 2012 #4

    SammyS

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    The result of this is no better than the original.
     
  6. Jan 28, 2012 #5
    don't use partial integral, first multiply the integrand ∫(√x/√(x−1))dx to √x/√x..

    Moderator note: I removed the subsequent work shown. Please let the OP try to work out the problem on his or her own.
     
    Last edited by a moderator: Jan 29, 2012
  7. Jan 29, 2012 #6
    ^

    Adding that extra sqrt(x) was clever. I seem to understand and accept with everything, but there is the part

    "divide 2x-1 to x"?

    U mean I calculate u = 2x-1 [itex]\Leftrightarrow[/itex] x = (1/2)(u+1)?
    which is then x dx = (1/2)(u+1) du?


    I'm sorry I might seem like totally idiot, but until university, sign (dx/du) was totally unknown to me so I'm not familiar with it and don't know how it exactly behaves.

    To HallsOfIvy, thanks for the tip.
     
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