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'simple' integration question

  1. Jun 29, 2006 #1

    im trying to integrate the following:

    e^(2x)sqrt(e^x - 1)

    if only it were e^x at the front - it'd be so much easier!!

    anyway, letting u=e^x-1 and doing everything i eventually get the following (and this is where i get stuck):

    [ u^(5/2)+2u^(3/2)+u^(1/2) ] [ 1/(u+1) ] du

    this is correct because, according to the computer, it simplifies to u^(3/2) + u^(1/2), which then gives the right answer.

    so my question, really, is: how do we simplify [ u^(5/2)+2u^(3/2)+u^(1/2) ] [ 1/(u+1) ] ??

    or am i approaching this the wrong way (which i suspect because it's a multiple choice question)??

  2. jcsd
  3. Jun 29, 2006 #2


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    Looks like a nice substitution...

    after which I end up with:

    [tex]\int e^{2x}\sqrt{u}\frac{du}{e^x}=\int e^{x}\sqrt{u} du=\int (u+1)\sqrt{u} du[/tex]

    which is what your computer says.

    (for your simplification: take a factor of [tex]u^{1/2}[/tex] out and factorise...)
    Last edited: Jun 29, 2006
  4. Jun 29, 2006 #3
    [tex] (\sqrt u)^5 + (\sqrt u)^3 + \sqrt u= \sqrt u \left(u^2 + 2u + 1)\right[/tex]
  5. Jun 29, 2006 #4


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    [itex] (u^{5/2} + 2 u^{3/2} + u^{1/2}) = u^{1/2} ( u^2+ 2 u + 1) = u^{1/2} (u+1)^2 [/itex]. That's all there was to it!

  6. Jun 29, 2006 #5


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    As that's sort of a lucky break peculiar to this example, a better substitution for this general type of problem might be [itex]u=\sqrt{e^x-1}[/itex]. Then [itex]e^x=u^2+1[/itex] and [itex]e^x dx = 2u du [/itex].
  7. Jun 29, 2006 #6
    oh, thanks a lot guys.

    it's so much quicker and easier if i just cancel (u+1)^2 and 1/(u+1), leaving just (u+1) instead of expanding and then trying to simplify...

    seems like i approached this q in the most inefficient of ways -- i even had x= ln (u+1) lol... i had like e^[ln(u+1)] = u+1... dont know what i was thinking...

    oh, by the way, do any of you guys know where i could find a good summary of all of the different types of antidifferation (ie like a compendium with an example of every particular type of question or someting)?

  8. Jun 30, 2006 #7


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    You'll usually find "antidifferentiation" :tongue: in the front of most calculus books - inside covers.

    Or Abramowitz and Stegun: Handbook of Mathematical Functions is the Giant Haystacks of the formulae world.
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