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Difficult Indefinite Integral (substitution problem?)

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\int (\frac{x}{\sqrt{x+8}}) dx [/itex]

    3. The attempt at a solution
    I gotta be honest, I don't even really know where to start with this problem. So bear with me as I take a wild stab in the dark. This section was substitution problems, ...

    [itex]\frac{x}{\sqrt{x+8}} = \sqrt{ \frac{x^2}{x+8} } =




    [/itex]
     
  2. jcsd
  3. Sep 4, 2011 #2

    rock.freak667

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    Try putting u = x+8, that should help you along.
     
  4. Sep 4, 2011 #3

    Ray Vickson

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    Even better: try putting sqrt(x+8) = u.

    RGV
     
  5. Sep 4, 2011 #4
    [itex]

    u = \sqrt{x+8} \rightarrow du = (1) \frac{1}{2 \sqrt{x+8}} dx

    [/itex]

    I don't think that helps...

    [itex]

    u = x+8 \rightarrow du = 1 dx

    [/itex]

    I don't think that helps, either. Am I missing something obvious?
     
  6. Sep 4, 2011 #5

    gb7nash

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    Yes. There's an x on the numerator of the integrand. If u = x + 8, x = ...
     
  7. Sep 4, 2011 #6
    Like stated above, u=x+8 makes the integral much simpler to solve.
     
  8. Sep 4, 2011 #7

    dynamicsolo

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    If you apply Mr. Vickson's method, you need to write [itex]u^{2} = x + 8 [/itex], which leads to [itex] 2u du = dx [/itex]; in the numerator, you then replace the remaining x in the numerator with u2 - 8 . More than one approach works for this integral.
     
  9. Sep 4, 2011 #8
    I ended up going,

    [itex]
    u = x + 8
    [/itex]

    [itex]
    x = u - 8
    [/itex]

    [itex]
    \int [u-8]/[sqrt(u)] du
    [/itex]

    [itex]
    \int (u-8)u^{-1/2} du
    [/itex]

    [itex]
    \int u^{1/2} - 8u^{-1/2} du =
    [/itex]

    [itex]
    c + \frac{2}{3} u^{3/2} - 16 u^{1/2}
    [/itex]

    [itex]
    c + \frac{2}{3} (8+x)^{3/2} - 16 (8+x)^{1/2}
    [/itex]

    Thanks everybody!
     
    Last edited: Sep 4, 2011
  10. Sep 4, 2011 #9

    gb7nash

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    Your method is fine, but look at this again.
     
  11. Sep 4, 2011 #10
    gb7nash, I went back and fixed that problem.

    Hey guys, thanks to all of you! You've been a great help.
     
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