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Math Help - integration by substitution

  1. #1
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    integration by substitution

    \int\frac{1}{e^x+1}dx (let u=e^x+1)

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

    =\int\frac{1}{u^2-u}du

    =\int\frac{1}{(u-\frac{1}{2})^2}-\frac{1}{4}du

    =ln(\frac{u-1}{u})+c

    =ln(\frac{e^x}{e^x+1})+c
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: integration by substitution

    You can wright the numerator as:
    \int \frac{1}{e^x+1}dx= \int \frac{(e^x+1)-(e^x)}{e^x+1}dx=\int \frac{e^x+1}{e^x+1}dx - \int \frac{e^xdx}{e^x+1}
    Now use the substitution t=e^x+1 for the second integral.
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  3. #3
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    Re: integration by substitution

    Quote Originally Posted by Punch View Post
    \int\frac{1}{e^x+1}dx (let u=e^x+1)

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

    =\int\frac{1}{u^2-u}du

    =\int\frac{1}{(u-\frac{1}{2})^2}-\frac{1}{4}du

    =ln(\frac{u-1}{u})+c

    =ln(\frac{e^x}{e^x+1})+c
    This is the same mistake you made in the last question. If there is a u^2 term in the denominator, the integral is NOT a logarithm unless there is a u term in the numerator.

    I would start by multiplying top and bottom by e^x...
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  4. #4
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    Re: integration by substitution

    Here's another way

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

    and let u = 1 + e^{-x}
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