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Integral using series

  1. Sep 20, 2006 #1

    G01

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    [tex]\int \frac{x-\tan^{-1} x}{x^3}[/tex]

    I know the series form of tan-1 x = [tex] \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} [/tex]

    I know I need to subtract the x from that series and divide the x cubed form that series but i can't seem to be able to right the resulting series in a general form, any hints?

    I thought it would be: [tex] -1/3x + \sum_{n=2}^{\infty} -x^{2n+2}/(2n+1)(2n+2)[/tex]. But this isnt it. Any Help?
     
  2. jcsd
  3. Sep 20, 2006 #2
    I don't see how you arrived at that series. What you may want to consider doing also, for simplicity's sake, is to separate the integral into one that can be easily evaluated and one that should be done by series. Redo your solution paying particular attention to integrating -arctan(x)/x^3
     
  4. Sep 21, 2006 #3

    StatusX

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    Dividing by x^3 just reduces the power of x by 3, ie, x^n/x^3=x^(n-3). Your answer is close.
     
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