Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Basic integration

  1. Apr 16, 2006 #1
    i've been having trouble solving this one for highschool math:

    it's under the substitution section


    [e^(x^0.5) ] / (x^0.5)

    do you guys think you can help me with this one?

  2. jcsd
  3. Apr 16, 2006 #2
    Have you tried a substitution?? There are only a couple of choices to choose from, and one is really obvious.

    And the LaTex for this equation is

    [tex] \int \frac{e^{x^{\frac{1}{2}}}}{x^{\frac{1}{2}}}dx [/tex]

    you can click on the equation to see the LaTex code.
    Last edited: Apr 16, 2006
  4. Apr 16, 2006 #3
    Actually, once you have more experience integrating exponential functions, you will realize a substitution is not really necessary.
  5. Apr 16, 2006 #4
    i'm still lost, and this is supposed to be a routine question :(

    i think there is a property of e that i am not familiar with that might be holding me back. any hints?

    i've looked at two ways, neither of which seem to be any good.

    1. tried substituting u for x^.5, which seemed to do no good as it didnt offer a du/dx.
    2. tried substiuting u for e^x^.5, using the idea that ln u = x^.5 for the denominator. again, this didn't offer me much help.

    the answer for this one is 2e^x^.5, and working backwards didn't seem to give me any more ideas. sorry if i'm missing something blatant :(
    Last edited: Apr 16, 2006
  6. Apr 16, 2006 #5
    To solve the problem via substitution, what substitution should you try?

    To solve it in a more direct manner, try differentiating the numerator. What do you realize?
  7. Apr 16, 2006 #6
    Let [tex] u=x^\frac{1}{2} [/tex],
    so [tex] x=u^2 ;dx=2udu [/tex]
    substitutes them into your integral, it become:
    [tex] \int \frac{e^u}{u}2udu [/tex]
    Last edited: Apr 16, 2006
  8. Apr 16, 2006 #7
    ah. :( after all the fuss about that root of x... thanks very much for all the help
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook