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Thread: Stuck integrating with "e"

  1. #1
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    Stuck integrating with "e"

    I'm stuck trying to solve this integral with "e". I am using the substitution method and made a sub, but cannot figure out how to apply it to this particular integral due to the sign of the exponent.
    Attached Thumbnails Attached Thumbnails Stuck integrating with "e"-stuckstep4.png  
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  2. #2
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    Can't you say the following,

    $\displaystyle \int (\frac{1}{4}e^{7x} + \frac{1}{e^{7x}})dx$

    EDIT: Nevermind, this is incorrect, I'll edit this post if I find something else.

    Harish beat me to it!
    Last edited by jegues; Oct 9th 2010 at 10:50 AM.
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  3. #3
    MHF Contributor harish21's Avatar
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    $\displaystyle 4e^{-7x}+e^{7x} = \dfrac{4}{e^{7x}}+e^{7x}= \dfrac{4+e^{2(7x)}}{e^{7x}}$

    $\displaystyle \int \dfrac{dx}{4e^{-7x}+e^{7x}} = \int \dfrac{e^{7x}}{4+e^{2(7x)}} dx$

    now let $\displaystyle u = e^{7x} $ and complete
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  4. #4
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    Stuck integrating with "e"-img_0002.jpg
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  5. #5
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    Thank you this helped me solve it and was able to verify with Pandevil1990's scan.
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  6. #6
    Junior Member Hardwork's Avatar
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    Let $\displaystyle \displaystyle u = e^{7x}$, then $\displaystyle \displaystyle \int \dfrac{dx}{4e^{-7x}+e^{7x}}\;{dx}$ $\displaystyle \displaystyle = \frac{1}{7}\int\frac{dx}{(\frac{4}{u}+u)u}\;{du} = \frac{1}{7}\int\frac{1}{4+u^2}\;{du} $. Let $\displaystyle \displaystyle t = 2\tan{u}$ to get $\displaystyle \displaystyle I = \frac{1}{14}\arctan\left(\frac{e^{7x}}{2}\right)+k$.
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