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Help finding the indefinite integral

  1. Jul 13, 2010 #1
    1/(4+3x^2) ?

    I'm trying to use this integration formula to solve for this

    du
    ----------
    a^2+u^2

    I know my a would be 2, but what would my u be? I tried u = 3x and du=3dx and du/3=dx

    but (3x)^2 would be 9x^2
     
  2. jcsd
  3. Jul 13, 2010 #2

    Mark44

    Staff: Mentor

    You want u2 = 3x2, so what should u be?
     
  4. Jul 13, 2010 #3
    sqrt(3x^2)

    but then du would be 1/2(3x^2)^(-1/2)*6x dx

    which simplifies to

    6x dx
    -------------
    2*sqrt(3x^2)
     
  5. Jul 13, 2010 #4

    Char. Limit

    User Avatar
    Gold Member

    Don't forget to use the property of...

    [tex]\sqrt{a b} = \sqrt{a}\sqrt{b}[/tex]
     
  6. Jul 13, 2010 #5

    Mark44

    Staff: Mentor

    How about u = sqrt(3)x, which makes du = sqrt(3)dx?
     
  7. Jul 13, 2010 #6
    Mark

    if U is sqrt(3x) which can be written as 3x^1/2

    when I find the du, it becomes

    1/2*(3x)^(-1/2)*3 dx

    which is written as

    3 dx
    ------
    2*sqrt(3x)

    I'm trying to just get du to equal dx so I can use the formula. Should I just multiply the denominator over and divide by 3?
     
  8. Jul 13, 2010 #7

    Char. Limit

    User Avatar
    Gold Member

    Chan, Mark said to use...

    [tex]u = \sqrt{3} x[/tex]

    You're using there...

    [tex]u = \sqrt{3x}[/tex]

    Note the difference.
     
  9. Jul 13, 2010 #8
    ooooh, sorry.

    When I get du = sqrt(3)dx, I just got dx by itself by dividing.

    It was plug and chug from there

    1
    ------- arctan(sqrt(3)*x) / 2 + C
    2*sqrt(3)
     
  10. Jul 13, 2010 #9

    Mark44

    Staff: Mentor

    I didn't check, but it looks like you might have an extra 2. You can check by differenting your answer.
     
  11. Jul 13, 2010 #10
    Hmm, i think i followed the template

    1/a arctan (u/a) + C

    1 1
    -- * ----- arctan(sqrt(3)*x/2) + C
    2 sqrt(3)
     
  12. Jul 13, 2010 #11

    Mark44

    Staff: Mentor

    That's better. What you had before was different.
    1
    ------- arctan(sqrt(3)*x) / 2 + C
    2*sqrt(3) .
     
  13. Jul 13, 2010 #12
    Are there any tricks to find out the U's?
     
  14. Jul 13, 2010 #13

    Mark44

    Staff: Mentor

    With some practice it's pretty straightforward. You were trying to match your integral
    [tex]\int \frac{dx}{4 + 3x^2}[/tex]
    to an entry in a table of integrals like this:
    [tex]\int \frac{dx}{a^2 + u^2}[/tex]

    You were able to figure out that a = 2, but stumbled with you substitution for u. In this case you want u2 = 3x2, so u = sqrt(3)*x. I can't think of any tips or tricks that would be better than getting practice.
     
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