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Indefinite integral involving arctan and ln

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data

    ∫ 1/x arctan (lnx) dx


    2. Relevant equations



    3. The attempt at a solution
    1.U substitution. SO u = ln x, du= 1/x dx
    ∫ arctan u du

    2.by parts: u = arctan u du = 1/ 1+u^2
    v = 1 dv = du

    3. uv - ∫vdu = artan u - ∫ 1/ 1+u^2
    = arctan u - artan u = 0

    Did I answer this right?
     
  2. jcsd
  3. Apr 19, 2012 #2

    Dick

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    If du=dv, then v=u, not v=1! I'd also suggest when you get to ∫ arctan u du and if you want to use u and v for the variables in the integration by parts you change that to ∫ arctan w dw. Otherwise the variable naming gets really confusing.
     
    Last edited: Apr 19, 2012
  4. Apr 19, 2012 #3
    Also, u = arctan(u) du doesn't make sense! If you already did u-substitution, use v and w for integration by parts so your variables don't get mixed up.
     
  5. Apr 19, 2012 #4

    Dick

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    I think applejack meant u=arctan(u), which also doesn't make sense, and is part of the variable naming problem.
     
  6. Apr 19, 2012 #5
    You're right, he did, the du part is when he took the derivative.

    I'm used to putting the four equations (u, dv, du, v) in a box shape in my scratch work so I always get confused when it is formatted poorly on websites. Regardless, in addition to changing the variable, there should have been a comma inserted there.
     
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