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Integral form

  1. Apr 23, 2005 #1
    i don't see why the following integral:[tex]\int\frac{dx}{\sqrt{x}(x + 1)}[/tex] uses the arctan formula. i know how to solve integrals. i just don't see why or how you can rewrite it using the arctan formula. Isn't the arctan formula used for integrals of the form: [tex]\int\frac{du}{a^2 + u^2}=\frac{1}{a}arctan\frac{u}{a}[/tex]
  2. jcsd
  3. Apr 23, 2005 #2
    [tex]\int \frac{dx}{\sqrt{x}(x+1)}[/tex]
    you have to use u-substitution.
    so, you have:
    [tex]u = \sqrt{x}~~~du=\frac{1}{2 \sqrt{x}[/tex]
    now, here's where it gets tricky:
    now, you can substitute back in.
    [tex]\int \frac{du}{u^2+1}[/tex]
    now, just integrate and plug u back in, and you are done.

    *edit* it isn't let me post du...well, i cant see it from my comp, but du = 1/2sqrt(x)dx
    Last edited: Apr 23, 2005
  4. Apr 23, 2005 #3
    thanks i see now.
  5. Apr 23, 2005 #4
    i just read your edit. is your solution incorrect?
  6. Apr 24, 2005 #5


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    He missed the 2 when differentiating [itex] x=u^{2} [/itex].

  7. Apr 24, 2005 #6
    [tex]u = \sqrt{x}[/tex]
    [tex]du = \frac{1}{2}x^{-\frac{1}{2}}dx[/tex]

    ??? What is wrong with that?
  8. Apr 24, 2005 #7


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    Nothing,just that u didn't take it into account in the post i was referring to.

  9. Apr 24, 2005 #8
    hehe, yea i left out the 2...so its half the integral when you work it out. this computer at work was frustrating me, so i forgot it. (my excuse :rofl: )
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