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Integrate 1/x2+a2.dx=1

  1. Jan 18, 2007 #1
    integratio 1/x2+a2.dx=1
    intervel0 to +infinity
    find value of 'a'
     
  2. jcsd
  3. Jan 18, 2007 #2
    Help Me Inthis Question
     
  4. Jan 18, 2007 #3

    Hootenanny

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    Hi there imran,

    I'm guessing that your integrand, is something along the lines of;

    [tex]\int^{\infty}_{0} \frac{1}{x^2} + a^2 \;dx = 1[/tex]

    Please note, that the rules of this forum prohibit us to provide help to someone who has not attempted there homework question before hand (assuming of course this is homework). Perhaps, if you posted your attempt, we could help you further...
     
  5. Jan 18, 2007 #4
    The a^2 factor must be on the denominator, otherwise the integral is divergent both horizontally at infinity and vertically at zero.

    Just clearing up the interpretation. Since the original poster has access to the original problem anyway, this doesn't constitute helping him. ;)

    --Stuart Anderson
     
  6. Jan 18, 2007 #5

    Gib Z

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    The Integral [tex]\int \frac{1}{x^2 + a^2} dx[/tex] looks very familiar to me :)

    Maybe some trig will help you out?
     
  7. Jan 19, 2007 #6

    Hootenanny

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    You are quite correct of course!
    And so it should :wink:
    Good suggestion :approve:
     
  8. Jan 19, 2007 #7
    find volume of solid y=x2-1 ,x=2 ,y=0 is relvived about y-asis
     
  9. Jan 19, 2007 #8

    HallsofIvy

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    If this is a new problem, start a new thread.

    Also, since it is obviously homework, post it in the homework thread and show what you have tried!
     
  10. Mar 9, 2007 #9
    plz give me good sites to refer for integration,its applications
     
  11. Mar 9, 2007 #10
    [tex]\[
    \int {\frac{1}
    {{x^2 + a^2 }}dx} = \frac{1}
    {a}\tan ^{ - 1} \frac{x}
    {a} + c
    \][/tex]

    Just in case.
     
  12. Mar 10, 2007 #11
    does the following integral even exist?
    [tex]\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx[/tex]
     
  13. Mar 10, 2007 #12

    cristo

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    Look at the above post:

     
  14. Mar 10, 2007 #13

    Gib Z

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    The question was actually with the a^2 also in the denominator, look at above posts. And that integral you stated only exists if a=0.
     
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