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Math Help - 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

    \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

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

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

    using the u substitutions

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

    then using partial fractions

    1/((u)(u-1)) = A/u + B/(u-1)<br />

    1 = A(u-1) + B(u)

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

    so A=-1 and  B = 1

    your new integral becomes \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; December 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 u=e^x.

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

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

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

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