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Math Help - Simple integration problem

  1. #1
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    Simple integration problem

    I'm having trouble integrating the following:

    <br />
\int \frac{1}{x-\sqrt{x}} dx<br />

    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:

    <br />
\int \frac{1}{x-\sqrt{x}} dx<br />

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

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

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

    Or alternatively

    set t = \sqrt{x}.

    End up with \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:

    <br />
\int \frac{1}{x-\sqrt{x}} dx<br />

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

    now let u = \sqrt{x}

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

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

    and thus you have:

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

    finish it now
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