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Thread: Integral problem

  1. #1
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    Integral problem

    d(x) = *IntegralSign(b^(-x/c) / (f + ax))dx

    a, b, c, f are constants

    I'm stumped. A little help
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  2. #2
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    Mmm... I dunno, I think it doesn't have an elementary primitive.
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  3. #3
    Eater of Worlds
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    I concur with K. It appears it's not doable by elementary means.
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  4. #4
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    I was thinking that if you use u = f+ax and dx = du/a

    you might find the solution. I end up with this answer after my math:

    ln |ax + f| + C
    ---------------
    ab^(x/c)

    I'm not sure if its correct.
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  5. #5
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    Quote Originally Posted by cetaamoxclob View Post
    d(x) = *IntegralSign(b^(-x/c) / (f + ax))dx

    a, b, c, f are constants

    I'm stumped. A little help
    Quote Originally Posted by cetaamoxclob View Post
    I was thinking that if you use u = f+ax and dx = du/a

    you might find the solution. I end up with this answer after my math:

    ln |ax + f| + C
    ---------------
    ab^(x/c)

    I'm not sure if its correct.
    The Mathematic site gives the answer in terms of the error function, which can only be approximated. But let's see what your substitution does.

    $\displaystyle \int \frac{b^{-x/c}}{f + ax}~dx$

    Let $\displaystyle u = f + ax \implies dx = \frac{1}{a}~du$

    So
    $\displaystyle \int \frac{b^{-x/c}}{f + ax}~dx = \int \frac{b^{-(f - u)/(ac)}}{u}~\frac{du}{a}$

    $\displaystyle = \int \frac{b^{-f/(ac)}b^{u/(ac)}}{u}$
    (which is pretty much the same form as you started with.)

    I can see no way to integrate this (barring some kind of approximation.)

    -Dan
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  6. #6
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    A friend of mine thinks the answer has something to do with some sort of infinite series.
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