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Trouble evaluating an integral

  1. Jan 15, 2009 #1
    Indefinite Integral(arctan(x/2) / (x^2 + 4) )dx

    (sorry, I can't get my word equation editor to paste it here)

    I have stared at this problem for a while, but I can't figure it out. I made the triangle that has tan = x/2, and that triangle has a hypotenuse of sqrt(4 + x^2). But that doesn't seem to get me anywhere. Also, I thought using a u-substitution by u = arctan(x/2) would work, because it would get rid of the arctan, but thats not leading me anywhere either.
    Could someone point me in the right direction?
     
  2. jcsd
  3. Jan 15, 2009 #2

    Dick

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    u=arctan(x/2) is correct approach. What does that make du?
     
  4. Jan 15, 2009 #3
    That makes du = 1/ (1 + (x^2/4)) dx
    But how does that cancel out the 4 + x^2 in the denominator?
    I get:
    Integral(u *(1 + (x^2/4)) / 4 + x^2)du
     
  5. Jan 15, 2009 #4

    Dick

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    (4+x^2)=4*(1+x^2/4). But you are also missing a factor of (1/2) in du. Don't forget the chain rule.
     
  6. Jan 15, 2009 #5
    Oh!!! That's kind of tricky. Thank you for your help, I appreciate it!
     
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