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Arctan integration question

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data
    ∫xarctanx dx


    2. Relevant equations
    d/dx (arctanx) = 1/(1 + x^2)
    ∫1/(1 + x^2) = arctanx
    ∫u dv = uv - ∫v du

    3. The attempt at a solution
    Hi everyone,

    Here's what I've done so far:

    Use integration by parts:
    u = arctanx
    du = 1/(1 + x^2) dx

    dv = x dx
    v = (1/2)x^2

    ∫u dv = uv - ∫v du
    = arctanx(1/2)x^2 - (1/2)∫x^2/(1 + x^2) dx

    But how do you integrate this new integral without going back into arctanx, in which case you'll be left with zero?

    Thanks for any help
     
  2. jcsd
  3. May 13, 2009 #2

    benorin

    User Avatar
    Homework Helper

    Long division. [tex]\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}[/tex]
     
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