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Thread: Integral

  1. #1
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    Integral

    Hello! I'm stuck on the following integral: integrate 1 / (1 + ae^x) dx with a = constant term > 0... Thank you!
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  2. #2
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    Re: Integral

    $\displaystyle \frac{1}{1+ae^x} = \frac{1+ae^x -ae^x}{1+ae^x} = \frac{1+ae^x}{1+ae^x} - \frac{ae^x}{1+ae^x} = 1 - \frac{ae^x}{1+ae^x}$

    can you integrate the last expression on your own?
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  3. #3
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    Re: Integral

    $\displaystyle \int 1/(1+ae^x) dx$ substitute $\displaystyle u = 1+ae^x$ and $\displaystyle du = ae^x dx $
    multiply integral by $\displaystyle ae^x/(ae^x)$, so your integral becomes

    $\displaystyle \int 1/(1+ae^x) * ae^x/(ae^x) dx$

    using the u substitutions

    $\displaystyle \int 1/((u*u-1)) * du$ note: $\displaystyle ae^x = u - 1$

    then using partial fractions

    $\displaystyle 1/((u)(u-1)) = A/u + B/(u-1)
    $

    $\displaystyle 1 = A(u-1) + B(u)$

    $\displaystyle 1 = (A+B)u - A $ -> $\displaystyle A+B = 0 $ and $\displaystyle -A= 1$

    so $\displaystyle A=-1$ and$\displaystyle B = 1$

    your new integral becomes $\displaystyle \int -1/u + 1/(u-1) du $

    you should be able to do the rest

    Edit: skeeter's way is much faster (didn't see his post before posting this)
    Last edited by Ahasueros; Dec 30th 2012 at 06:20 PM.
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  4. #4
    Member sbhatnagar's Avatar
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    Re: Integral

    You can substitute $\displaystyle u=e^x$.

    $\displaystyle \int \frac{1}{1+ae^x}dx = \int \frac{du}{(1+au)u}$

    $\displaystyle =\int \left( \frac{1}{u}-\frac{a}{1+au}\right)du$

    $\displaystyle =\ln u -\ln(1+au)+C$

    $\displaystyle =x-\ln(1+ae^x)+C$
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