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A tricky integral

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data
    I should find this integral:
    [tex]\int[/tex]b x(1/[tex]\Pi[/tex])(1/(1+x2)dx

    2. Relevant equations

    [tex]\int[/tex]1/(1+x2)dx = arctan(x)

    3. The attempt at a solution
    The Only thing I've succeeded in doing is to take the 1/[tex]\Pi[/tex] and put it in front of the integral like this:
    (1/[tex]\Pi[/tex])[tex]\int[/tex]b (x/(1+x2)dx
    And I know that the integral of 1/(1+x2) equals arctan(x) but how could that help me? Ive tried to use the equation
    [tex]\int[/tex]f(x)g(x)dx = F(x)g(x) - [tex]\int[/tex]F(x)g'(x)dx
    but I can't compute the integral of arctan(x).
    Could someone help me?
  2. jcsd
  3. Jan 14, 2009 #2


    Staff: Mentor

    Is this the integral?
    [tex]\frac{1}{\pi}\int_{-b}^b{\frac{x}{x^2 + 1}dx[/tex]

    If so, you can use a simple substitution, u = x^2 + 1, and du = 2xdx
  4. Jan 14, 2009 #3


    Staff: Mentor

    BTW, you should have posted this in the Calculus & Beyond forum. This problem clearly falls in that area.
  5. Jan 14, 2009 #4
    Seems kind of weird to be doing that integral from -b to b, whether or not b > 0 or < 0 you get into complex numbers.
  6. Jan 14, 2009 #5

    Gib Z

    User Avatar
    Homework Helper

    Perhaps the easiest way is to observe this function is odd, and look at the interval of integration.
  7. Jan 14, 2009 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    No, you don't. You are squaring not taking square roots. As Mark44 said, use the substitution u= x2+ 1. Or, even simpler, use Gib Z's suggestion. This is really a very simple integral.
  8. Jan 14, 2009 #7
    Oh duh, good point.
  9. Jan 14, 2009 #8
    Yeah, sorry, I got it myself pretty soon after posting this. It feels like the more I study maths, the more I forget..!
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