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Math Help - perform the integration

  1. #1
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    perform the integration

    Can someone show me the solution to this problem? I have the exam tomorrow

    the integral of e^-3x/12+e^-3x

    the answer is: -1/3ln|12+e^-3x|+c

    Thank you!!!!
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  2. #2
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    So integrate (e^-3x)/(12 + e^-3x)?

    What does our natural log rule say? u'/u = ln|u| + c, right? Like the integral of 1/x is ln|x|. You've got the same thing here. Only difference is, u' would equal -3e^-3x. Well, we have -1/3 of that. Fix this by throwing a -1/3 outside the integral sign.

    Then it's just -1/3 \int u'/u. Or -1/3ln|u| + c
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  3. #3
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    Quote Originally Posted by venikUSA View Post
    Can someone show me the solution to this problem? I have the exam tomorrow

    the integral of e^-3x/12+e^-3x

    the answer is: -1/3ln|12+e^-3x|+c

    Thank you!!!!
    e^(-3x)/[12+e^(-3x)]

    in future, use grouping symbols to make your expressions clear ... or learn how to use latex

    \int \frac{e^{-3x}}{12 + e^{-3x}} \, dx

    -\frac{1}{3} \int \frac{-3e^{-3x}}{12 + e^{-3x}} \, dx

    -\frac{1}{3} \ln(12 + e^{-3x}) + C


    btw, you don't need the absolute value in this particular case ... why?
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