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Thread: integral question

  1. #1
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    integral question

    $\displaystyle \int\frac{1}{1+\sqrt{2x}}$

    according to my book it says let $\displaystyle u=1+\sqrt{2x}$

    $\displaystyle du=\frac{1}{\sqrt{2x}}dx-->(u-1)du=dx$

    I do not understand how $\displaystyle du=\frac{1}{\sqrt{2x}}dx-->(u-1)du=dx$

    when i try to take the derivative of U i get: $\displaystyle du= \frac{d}{dx}[1+\sqrt{2x}]=\frac{2}{3}(2x)^\frac{3}{2}dx$

    Im having problems getting from here:$\displaystyle \int\frac{1}{1+\sqrt{2x}}$ to here $\displaystyle \int\frac{(u-1)}{u}du$


    Thanks
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  2. #2
    MHF Contributor Unknown008's Avatar
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    $\displaystyle du=\frac{1}{\sqrt{2x}}dx$

    This becomes:

    $\displaystyle \sqrt{2x} du=dx$

    What is $\displaystyle \sqrt{2x}$?

    $\displaystyle u = 1 + \sqrt{2x}$

    $\displaystyle u - 1 = \sqrt{2x}$

    Hence;

    $\displaystyle \sqrt{2x} du=dx$

    $\displaystyle (u-1)du=dx$
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  3. #3
    MHF Contributor Unknown008's Avatar
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    $\displaystyle du= \dfrac{d}{dx}[1+\sqrt{2x}]=\frac{1}{2}(2x)^{-\frac{1}{2}}dx$

    This is what you should do. You integrated it instead of differentiating it.
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  4. #4
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    Thats it.. chain rule... got it.. thanks a lot..
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