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Thread: Integration with u sub

  1. #1
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    Integration with u sub

    integrate (-4-4e^x)/(1-e^x) using u substitution where u=e^x

    I've done it a few ways and gotten...
    4ln(1-e^-x)+4ln(1-e^x)+c
    8ln(1-e^x)-4x+c
    4ln(e^x)+4ln(1-e^x)+4ln(e^x-1)+c


    neither of which are correct, any insight on what I'm doing wrong?
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  2. #2
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    Re: Integration with u sub

    Hello, Johngalt13!

    If you inserted absolute values, your second answer is correct.


    $\displaystyle \text{Integrate: }\:\int\frac{-4-4e^x}{1-e^x}\,dx\:\text{ using u-substitution: }u\,=\,e^x$

    I've done it a few ways and gotten...
    . . $\displaystyle 4\ln|1 - e^{-x}| + 4\ln|1 - e^x| + C$
    . . $\displaystyle 8\ln|1 - e^x| - 4x + C$ **
    . . $\displaystyle 4\ln(e^x) + 4\ln|1 - e^x| + 4\ln|e^x - 1| + C$

    I solved it like this . . .

    $\displaystyle \int\frac{-4-4e^x}{1-e^x}\,dx \;=\;-4\int\frac{1+e^x}{1-e^x}\,dx \;=\;4\int\frac{e^x+1}{e^x-1}\,dx$

    $\displaystyle \text{Let }u = e^x \quad\Rightarrow\quad x = \ln u \quad\Rightarrow\quad dx = \frac{du}{u}$

    $\displaystyle \text{Substitute: }\:4\int \frac{u+1}{u-1}\,\frac{du}{u} \;=\;4\int\frac{u+1}{u(u-1)}\,dx$

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

    . . . . . . . . . . . $\displaystyle =\;4\big(2\ln|u-1| - \ln|u|\big) + C$


    $\displaystyle \text{Back-substitute: }\:4\big(2\ln|e^x-1| - \ln|e^x|\big) + C \;=\; $

    . . . . . . . . . . $\displaystyle =\;4\big(2\ln|e^x-1| - x\big) + C$

    . . . . . . . . . . $\displaystyle =\;8\ln|e^x-1| - 4x + C$


    $\displaystyle \text{Note that: }\:|e^x - 1| \:=\:|1-e^x|$
    Thanks from Johngalt13
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