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Homework Help: Integral Help Please

  1. May 3, 2004 #1

    Zurtex

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    Got an integral that for some reason I can't do, very much annoying me because this is supposed to be the stuff I am good at. A nudge in the right direction would be great, here it is:

    [tex]\int \frac{(x+2)dx}{\sqrt{x^2+6x+4}}[/tex]

    I have attempted completing the square on the bottom to get [itex](x+3)^2-5[/itex] and then I tried the substitution [itex]u=x+3[/itex]. After messing about with it a bit I got an answer but it didn't look right and it was a lot different from the answer I got off an online integrator and its answer really did look better.

    Any help would be great thanks.
     
  2. jcsd
  3. May 3, 2004 #2

    arildno

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    Have you tried the substitution:
    u= sqrt(5)*Cosh(t)?
     
  4. May 3, 2004 #3

    Zurtex

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    Thanks, could someone please confirm this is the answer then:

    [tex]\sqrt{ \frac{x^2-6x+4}{5} } - \cosh^{-1}\left(\frac{x-3}{\sqrt{5}}\right) + C[/tex]
     
  5. May 3, 2004 #4

    arildno

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    Well, I got the -6x term to be +6x, and the x-3 to be x+3
     
  6. May 3, 2004 #5

    HallsofIvy

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    Actually, I like your first idea of substituting for x+3. If you let u= x+3, then
    x2+6x+ 4= (x+3)2- 5= u2- 5 and x+2= x+3-1= u-1 so the integral becomes
    [tex]\int{\frac{u-1}{\sqrt{u^2-5}}du[/tex]
    which we can separate into
    [tex]\int{\frac{udu}{\sqrt{u^2-5}}+ \int{\frac{du}{\sqrt{u^2-5}}[/tex]

    Now you can make the substitution v= u2-5 in the first integral and
    [tex]sec(\theta)= \sqrt{5}u [/tex] in the second.
     
    Last edited by a moderator: May 4, 2004
  7. May 3, 2004 #6

    Zurtex

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    kk thanks for the help and thanks for the other approach :smile:
     
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