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Integration using substitution

  1. Apr 21, 2014 #1
    1. $$\int \frac{1}{1+e^x}\,dx$$

    2. Relevant equations


    3. The attempt at a solution

    $$u=1+e^x$$ $$du=e^xdx$$
    $$\int \frac{1}{u}\frac{1}{e^x}\,du$$ $$\int \frac{1}{u}\frac{1}{u-1}\,du$$ $$\int \frac{1}{u(u-1)}\,du$$
    $$= \ln {|u^2-u|} = \ln {|(1+e^x)-(1+e^x)|} = \ln {|1+2e^x+e^{2x}-1-e^x|} = \ln {|e^2|}$$

    There is something obviously wrong with my answer but I don't know what. The answer key says $$x-\ln {(1+e^x)}+C$$ Any help is greatly appreciated!
  2. jcsd
  3. Apr 21, 2014 #2
    The item in red is wrong .

    Write 1/u(u-1) = 1/(u-1) - 1/u .
  4. Apr 21, 2014 #3


    Staff: Mentor

    To expand on what Tanya Sharma said,
    $$\int \frac{dx}{f(x)} \neq ln|f(x)| + C$$
    unless f(x) just happens to be x.
  5. Apr 21, 2014 #4
    or, to be more precise f(x) happens to be of the form ax+b ,where a and b are constants.
    Last edited: Apr 21, 2014
  6. Apr 21, 2014 #5
    Yes subbing u = 1+e^x would work, but for this question you do not need to figure out du. So
    ∫1/(1+e^x) dx

    so 1 = u-e^x

    ∫(u-e^x)/u dx

    from here you can see how it would work.
    You could also do it your way but you have to use partial fraction decomposition.
  7. Apr 21, 2014 #6
    A better approach is to do this:
    $$\int \frac{1}{1+e^{x}}\,dx=\int \frac{e^{-x}}{1+e^{-x}}\,dx$$
    Use ##e^{-x}=u## to get:
    $$\int \frac{-du}{1+u}\,du=-\ln|1+u|+C=-\ln|1+e^{-x}|+C$$
    Last edited: Apr 21, 2014
  8. Apr 21, 2014 #7


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    Gold Member

    Or another: $$\int \frac{1}{1+e^x}\,\text{d}x = \int \frac{1+e^x}{1+e^x}\,\text{d}x - \int \frac{e^x}{1+e^x}\,\text{d}x$$The second integral on the right can be evaluated by a simple substitution.
  9. Apr 21, 2014 #8


    Staff: Mentor

    No, that wouldn't work.
    $$\int \frac{dx}{ax + b} = \frac 1 a ln|ax + b| + C$$
    So ∫dx/f(x) ≠ ln|f(x)| + C. There's that pesky factor of 1/a.
  10. Apr 21, 2014 #9
    How was she able to ln the instead like that? I thought there was a factor of [itex]\frac{1}{a}[/itex] to deal with?
  11. Apr 21, 2014 #10


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    Staff Emeritus
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    ... ln the WHAT instead ...

    What part of Pranav-Arora's post don't you get?
  12. Apr 21, 2014 #11
    I'm assuming they did something like this?

    $$\int \frac{-du}{u+1} = -\frac{1}{1}\ln {|u+1|} +C$$
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