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  1. #1
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    exam question

    this is my minus 20 points
    feel free to answer:
    $\displaystyle \int\frac{e^{2x}+4}{e^{2x}-4}dx$
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  2. #2
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    You could have written it as:

    $\displaystyle \frac{2}{e^{x}-2}-\frac{2}{e^{x}+2}+1$

    and went from there.

    $\displaystyle \int\frac{2}{e^{x}-2}dx=[ln(e^{x}-2)-x]$

    $\displaystyle \int\frac{2}{e^{x}+2}dx=[x-ln(e^{x}+2)]$
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  3. #3
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    $\displaystyle \displaystyle \int\frac{e^{2x}+4}{e^{2x}-4}dx=\int\left(1+\frac{8}{e^{2x}-4}\right)dx=x+8\int\frac{1}{e^{2x}-4}dx$.
    Substitute: $\displaystyle \displaystyle e^{2x}=t\Rightarrow 2x=\ln t\Rightarrow dx=\frac{1}{2t}dt$.
    Then $\displaystyle \displaystyle 4\int\frac{dt}{t(t^2-4)}=-\ln |t|+\frac{1}{2}\ln |t+2|+\frac{1}{2}\ln |t-2|+C$.
    So the original integral is
    $\displaystyle \displaystyle -x+\frac{1}{2}\ln |e^{2x}-2|+\frac{1}{2}\ln (e^{2x}+2)+C$
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  4. #4
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    Hello, ^_^Engineer_Adam^_^!


    Another approach . . .


    $\displaystyle \int\frac{e^{2x}+4}{e^{2x}-4}dx$
    Let $\displaystyle e^{2x} = u\quad\Rightarrow\quad x = \frac{1}{2}\ln u\quad\Rightarrow\quad dx = \frac{du}{2u}$

    Substitute: .$\displaystyle \int\frac{u+4}{u-4}\left(\frac{du}{2u}\right) \;=\;\frac{1}{2}\int\frac{u+4}{u(u-4)}\,du$

    Partial Fractions: .$\displaystyle \frac{1}{2}\int\left(\frac{2}{u-4} - \frac{1}{u}\right)du\;=\;\frac{1}{2}\left[2\ln(u-4) - \ln u\right] + C
    $

    Back-substitute: .$\displaystyle \frac{1}{2}\left[2\ln(e^{2x} - 4) - \ln(e^{2x})\right] + C \;=\;\frac{1}{2}\left[2\ln(e^{2x} - 4) - 2x\right] + C$

    . . . . . . . . . . $\displaystyle = \;\ln(e^{2x}-4) - x + C$

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  5. #5
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    or in the integral $\displaystyle \int \frac{4}{e^{2x}-4}dx$ you could multiply both numerator and denominator by
    $\displaystyle \frac{e^{-2x}}{e^{-2x}}$

    btw thanks for the help!
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