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Integral help

  1. Feb 1, 2008 #1
    how do i integrate 1/(x^2 +4)?

    please help
     
  2. jcsd
  3. Feb 1, 2008 #2
    What's the derivative of arctan?
     
  4. Feb 1, 2008 #3
    1/1+x^2
     
  5. Feb 1, 2008 #4
    See how that might be helpful?
     
  6. Feb 1, 2008 #5

    rock.freak667

    User Avatar
    Homework Helper

    [tex]\int \frac{1}{x^2 +4} dx[/tex]

    [tex] = \int \frac{1}{x^2 +(2)^2} dx[/tex]


    try x=2tan[itex]\theta[/itex]
     
  7. Feb 1, 2008 #6
    hmm lemem see...1 second
     
  8. Feb 1, 2008 #7
    Exactly.. so now from your function, you take 4 common, you get:

    [tex]
    \frac{1}{4}\int\frac{1}{({\frac{x}{2}})^2 + 1}dx
    [/tex]

    Now, if you take [itex]\frac{x}{2} = y[/itex].. you can solve this integral.. get a hint?

    Once you have done this, it would be helpful for you to remember the formula for a general case as in [tex]\int\frac{1}{a^2 + x^2}dx[/tex]
     
  9. Feb 1, 2008 #8
    Yeah rohan's post is what I was think too..
     
  10. Feb 1, 2008 #9
    oh i see, thanks

    and quick response from everyone =)
     
  11. Feb 2, 2008 #10
    [tex]\int\frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C[/tex]
    [tex]\int\frac{1}{x^2+2^2}dx=\frac{1}{2}\arctan\frac{x}{2}+C[/tex]
     
  12. Feb 2, 2008 #11
    No need to post the solution he all ready figured it out...
     
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