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Math Help - integration problem (partial fractions)

  1. #1
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    integration problem (partial fractions)

    i tried this and really have no idea how to approach this particular question:
    apparently, i have to first substitute e^x=u and therefore du=e^xdx which gives du/u=dx but when i replace these values, i find that A/(u+2) + B/(u+3)du/u...From there, i find that A=-2 and B=-15 but when i substitute these values, i find (-2ln(e^x+2)-15ln(e^x+3))/e^x. The final answer is supposed to be: -6x+ln(2+e^x)+5ln(3+e^x). Where am i going wrong?
    Thanks.
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  2. #2
    MHF Contributor matheagle's Avatar
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    You can break this into two separate integrals.
    The first one is easy...
    -17\int{e^xdx\over e^{2x}+5e^x+6}.
    Let u=e^x and you have -17\int{du\over u^2+5u+6}.
    Note that u^2+5u+6=(u+3)(u+2) and that's done.
    I can do the secon one, but there must be an easier way.

    -36\int{dx\over e^{2x}+5e^x+6} can be done by first multiplying to and bottom by e^{-x}
    and then letting u=e^x again.
    But there should be a better way, something to do with the 17 and 36.
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  3. #3
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    consider, for a moment, that you simply did partial fraction decomposition on the integrand, which leaves you needing to solve

    \int \frac{-2}{e^x+2} \, dx + \int \frac{-15}{e^x + 3} \,dx

    Now consider

    \int \frac{1}{e^x+1} \,dx

    = \int \frac{1 + e^x - e^x}{1+e^x} \,dx

    = \int 1 - \frac{e^x}{1+e^x} \,dx

    = x - \ln(1+e^x)

    Is this hint enough to get you going now?
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  4. #4
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Stonehambey View Post
    consider, for a moment, that you simply did partial fraction decomposition on the integrand, which leaves you needing to solve

    \int \frac{-2}{e^x+2} \, dx + \int \frac{-15}{e^x + 3} \,dx

    Now consider

    \int \frac{1}{e^x+1} \,dx

    = \int \frac{1 + e^x - e^x}{1+e^x} \,dx

    = \int 1 - \frac{e^x}{1+e^x} \,dx

    = x - \ln(1+e^x)

    Is this hint enough to get you going now?

    I like the trick where you multiply the numerator and denominator of

    \int \frac{dx}{e^x+1} by e^{-x} giving you \int \frac{e^{-x}dx}{1+e^{-x}} .

    Then let u=1+e^{-x}.
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