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Integral of x/sqrt(x + 2)

  1. Jul 5, 2013 #1
    [itex]\int\frac{x}{\sqrt{x + 2}}dx[/itex]

    We are still using substation as our method of solving integrals. I've rationalized the denominator, but that doesn't seem to help a whole lot. Any value for u I've picked so far hasn't worked. I've looked up the solution online, and I know it's not a trig integral. Any small hint would help.
     
  2. jcsd
  3. Jul 5, 2013 #2

    CompuChip

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    Once you recognise that
    ##\frac{1}{\sqrt{x + 2}} \propto \frac{d}{dx} \sqrt{x + 2}##
    you could try integration by parts.
     
  4. Jul 5, 2013 #3

    LCKurtz

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    Or you could try ##x+2=u^2##.
     
  5. Jul 5, 2013 #4

    ehild

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    Or u=x+2 :biggrin:

    ehild
     
  6. Jul 5, 2013 #5

    vanhees71

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    ...or you use a little trick:

    [tex]
    \int \mathrm{d} x \frac{x}{\sqrt{x+2}}=\int \mathrm{d} x \frac{x+2-2}{\sqrt{x+2}} = \int \mathrm{d} x \left [(x+2)^{1/2}-2 (x+2)^{-1/2} \right ]=\frac{2}{3} (x+2)^{3/2} - 4 (x+2)^{1/2}+\text{const}.[/tex]
     
  7. Jul 5, 2013 #6
    Wow, sweet moves bro... I like that trick. Thanks a million.
     
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