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Integral of x arctan x dx

  1. Mar 25, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex] \int x \arctan x \, dx [/tex]

    3. The attempt at a solution
    By parts,
    [tex] u = \arctan x[/tex]
    [tex] dv = x dx[/tex]
    [tex] du = \frac{dx}{x^2+1}[/tex]
    [tex]v = \frac{x^2}{2} [/tex]

    [tex] \int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{1}{2} \int \frac{x^2}{x^2+1} \, dx [/tex]

    Again...by parts

    [tex] u = x^2 [/tex]
    [tex] dv = \frac{dx}{x^2+1} [/tex]
    [tex] du = 2x dx [/tex]
    [tex] v = arc tan x [/tex]

    [tex]\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{x^2}{2}\arctan x - \int x \arctan x \, dx [/tex]
    I back to the beginning, what did wrogn?

    [tex]\int x \arctan x \, dx = - \int x \arctan x \, dx [/tex]
     
    Last edited: Mar 25, 2007
  2. jcsd
  3. Mar 25, 2007 #2
    [tex]\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{x^2}{2}\arctan x - \int x \arctan x \, dx [/tex]

    Add [tex]\int x \arctan x \, dx[/tex] to both sides, then solve for the integral, assuming your work is correct.
     
  4. Mar 25, 2007 #3
    you mean like this? is the same, i back to the beginign

    [tex]\int x \arctan x \, dx +\int x \arctan x \, dx = \frac{x^2}{2}\arctan x - \frac{x^2}{2}\arctan x - \int x \arctan x \, dx +\int x \arctan x \, dx[/tex]

    [tex]2\int x \arctan x \, dx = 0[/tex]
     
  5. Mar 26, 2007 #4
    Why use 'by parts' again? It would easier if you just add and subtract 1 from the numerator
     
  6. Mar 26, 2007 #5
    why not try the substitution u=x^2+1 in that second integral...
     
  7. Mar 26, 2007 #6
    for the integral x²/(x²+1)
    you can rewrite it as (x² + 1 - 1)/(x²+1) => 1 - 1/(x²+1)
     
  8. Mar 26, 2007 #7

    Gib Z

    User Avatar
    Homework Helper

    umm hmm, that leaves a nice (x - arctan x) for you there.
     
  9. Dec 1, 2008 #8
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