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Can this be integrated?

  1. May 3, 2014 #1
    1. The problem statement, all variables and given/known data

    ∫( log(1 - x) / (x + 1), x )

    I have tried integrating this and had no luck.

    2. Relevant equations

    As above

    3. The attempt at a solution

    By parts

    u = 1/(x+1) , du = ln(x+1)dx
    dv = log(1-x) dx , v = (x-1) * ( log(1-x) - 1 )

    uv - int(v du) = 1/(x+1) * (x-1) * ( log(1-x) - 1 ) - ∫( (x-1)*(log(1-x) - 1) * log(x+1) dx )

    Gets worse as I continue
     
  2. jcsd
  3. May 3, 2014 #2

    Dick

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    It's not an elementary integral you can compute that way. I think you need nonelementary functions like polylogarithms.
     
  4. May 4, 2014 #3
    I don't understand what this means: "∫( log(1 - x) / (x + 1), x )" .... why is there a comma? where is dx ?
    d(ln(y))/dy = 1/y is true.
    d(1/y) = ln(y) is NOT.
    if x >= 1, then log(1-x) is undefined. so why do you have an indefinite integral? not to mention 1/(x+1) when x= -1 ?
    But I don't see how to solve it...I'd try trig functions...but its been way too long...Dick may be (and probably is, for all I know) exactly correct - its not elementary, IDK.
     
  5. May 4, 2014 #4

    ehild

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    Is it not ##\int{\log\left(\frac{1-x}{x+1}\right)dx}##?

    ehild
     
  6. May 4, 2014 #5

    LCKurtz

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    I interpret it as ##\int \frac{\log(1-x)}{1+x}~dx##
     
  7. May 4, 2014 #6
    Isn't that just separation?

    Looking how OP solved, this seems more correct

    Even I tried it I couldn't do it.
    Polylogarithms was correct when I tried to do it with Wolfram Integrator.
    http://mathworld.wolfram.com/Polylogarithm.html
     
  8. May 4, 2014 #7
  9. May 4, 2014 #8
    Are you sure that you were asked to calculate the anti derivative instead of some definite integral?
     
  10. May 4, 2014 #9
    It's a definite integral, so I've always calculated this by obtaining the anti-derivative and subbing in the limits of integration.
     
  11. May 4, 2014 #10
    Then you should have posted the definite integral. You can not evaluate every definite integral by obtaining the anti-derivative, there is a different way to handle them. In any case, can you post the definite integral you are assigned to solve?
     
  12. May 4, 2014 #11
    [tex]\int_0^x \frac{1}{b + 2ts}\textrm{ln}\left(\frac{b+2t(L-s)}{b}\right)ds[/tex]

    That's one of the integrals I extracted from the Peano Baker series when attempting to solve a system of second order ODE with variable coefficients
     
    Last edited by a moderator: May 4, 2014
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