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Thread: Simple integration problem

  1. #1
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    Simple integration problem

    I'm having trouble integrating the following:

    $\displaystyle
    \int \frac{1}{x-\sqrt{x}} dx
    $

    I can tell that I'm going to have to do a few substitutions... but I have no idea where.
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  2. #2
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Positronic View Post
    I'm having trouble integrating the following:

    $\displaystyle
    \int \frac{1}{x-\sqrt{x}} dx
    $

    I can tell that I'm going to have to do a few substitutions... but I have no idea where.
    $\displaystyle
    \int \frac{1}{x-\sqrt{x}} dx = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)} dx
    $

    Then use partial fractions... (Haven't checked if it works though)

    edit: Don't think this will help actually lol.

    Or alternatively

    set $\displaystyle t = \sqrt{x}$.

    End up with $\displaystyle \int \frac{2t}{t^2 - t}dt$ which you can simplify and will be easier to solve.


    EDIT 2: Got t^2 and t mixed up. Above is correct.
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Positronic View Post
    I'm having trouble integrating the following:

    $\displaystyle
    \int \frac{1}{x-\sqrt{x}} dx
    $

    I can tell that I'm going to have to do a few substitutions... but I have no idea where.
    $\displaystyle \int \frac{1}{x-\sqrt{x}} dx = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)}$

    now let $\displaystyle u = \sqrt{x}$

    then $\displaystyle du = \frac{1}{2\sqrt{x}} dx$

    and $\displaystyle dx = 2\sqrt{x} du = 2u \mbox{du}$

    and thus you have:

    $\displaystyle \int \frac{2u}{u(u-1)}du = \int \frac{2}{u-1}du$
    .
    .
    .

    finish it now
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