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Integral of ln(x)/(x+1) from 1 ot 0

  1. May 17, 2010 #1
    what is it's solvation.I tried to solve it by parts but i found undefinable term(from ln(0)).Please give me an advice.Thanks.
  2. jcsd
  3. May 17, 2010 #2


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    This integral doesn't have a primitive in terms of elementary functions.

    Edit: didn't notice it was a definite integral.
  4. May 17, 2010 #3


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    That should be a clue that you've forgotten about improper integrals....
  5. May 20, 2010 #4
    If i integrete this integral by parts


    ln(x+1)=u ,1/(x+1)dx=du and 1/xdx=dv, lnx=v

    [tex]\int_{0}^{1}\frac{ln(x+1)}{x}dx=ln(x+1)lnx\mid_{0}^{1}-\int_{0}^{1} \frac{lnx}{x+1}dx[/tex]

    This time integral of ln(x+1)/x from 0 to 1 is not integratable but answer of this is (pi^2)/6.
  6. May 20, 2010 #5


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    You can integrate it just fine, however there does not exist a primitive in terms of elementary functions. You can evaluate this integral by using the power series of ln(x+1).
  7. May 20, 2010 #6
    The message #4 is in contradiction with the title of the thread…
    Which integral is it ?
  8. May 20, 2010 #7

    Gib Z

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    No it isn't. The post showed that each integral yields the other. And as Cyosis mentioned, using the Taylor Series for ln(1+x) centered at x=0, we can get a series representation of the integral, and it so happens in this case it simplifies to a simple combination of well known constants.
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