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Integral of 1/(x^2*lnx) !

  1. Feb 20, 2012 #1
    ∫dx/(x^2*lnx)

    What I`ve seen on the web but I don`t think is right:
    u= lnx *** what we have here isn't lnx but (lnx)^-1... This is why I doubt that's the right solution
    du = dx/x
    dv = dx/x^2
    v = -1/x

    =-lnx/x + ∫dx/x^2
    =-lnx/x - 1/x + C

    Let me know if it is correct, thanks!
     
  2. jcsd
  3. Feb 20, 2012 #2
    I didn't check if it is correct myself, but you never really need someone to check if an integral (like this) is correct. Just differentiate it.
     
  4. Feb 20, 2012 #3

    Ray Vickson

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    Maple 14 gets a non-elementary function for this integral; that is, it cannot be done in terms of elementary functions of the type you have used.

    RGV
     
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