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Math Help - Help with Natural exponential integrals

  1. #1
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    Help with Natural exponential integrals

    Hi all!
    im stuck again with this difficult integral

    \int \frac{dx}{1+e^x}

    can u help with what to substitute
    ??

    and we're not yet in integration by parts
    Last edited by ^_^Engineer_Adam^_^; June 25th 2007 at 05:16 AM.
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Hi all!
    im stuck again with this difficult integral

    \int \frac{dx}{1+e^x}

    can u help with what to substitute
    ??

    and we're not yet in integration by parts
    Try the substitution u=e^x, then the integral becomes:

    <br />
\int \frac{1}{u(1+u)} du = \int \frac{1}{u}-\frac{1}{1+u}du<br />

    RonL
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  3. #3
    MHF Contributor

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    Here is a second way:
    \int {\frac{{dx}}{{1 + e^x }}}  = \int {\frac{{e^{ - x} dx}}{{e^{ - x}  + 1}} =  - \ln \left( {e^{ - x}  + 1} \right)}.
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  4. #4
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    Quote Originally Posted by Plato View Post
    Here is a second way:
    \int {\frac{{dx}}{{1 + e^x }}} = \int {\frac{{e^{ - x} dx}}{{e^{ - x} + 1}} = - \ln \left( {e^{ - x} + 1} \right)}.
    I think you forgot something
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  5. #5
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    can u help with what to substitute
    ??

    and we're not yet in integration by parts
    I suggest when you post a problem, expose your solution or part of it to see the progress and possibles mistakes that you could have.

    Regards
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Hi all!
    im stuck again with this difficult integral

    \int \frac{dx}{1+e^x}

    can u help with what to substitute
    ??

    and we're not yet in integration by parts
    Yet another possibility:

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

    = \int \left( 1 - \frac {e^x }{1 + e^x} \right)~dx

    = x - \ln |1 + e^x| + C

    The possibilities are endless
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    Yet another possibility:

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

    = \int \left( 1 - \frac {e^x }{1 + e^x} \right)~dx

    = x - \ln |1 + e^x| + C

    The possibilities are endless
    I like this solution.
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  8. #8
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    Hello, ^_^Engineer_Adam^_^!

    This is one of my favorite integrals.
    I posted a long discourse on it on SOSMath in 2003.


    \int \frac{dx}{1+e^x}
    (1) Plato had the best method: .multiply top and bottom by e^{-x}
    His answer: . -\ln(e^{-x} +1) + C



    (2) CaptainBlack also had a good approach: Partial Fractions.

    He had: . \int\left(\frac{1}{u} - \frac{1}{u+1}\right)\,du \;=\;\ln(u) - \ln(u + 1) + C

    Back-substitute: . \ln(e^x) - \ln(e^x + 1) + C \;=\;x - \ln(e^x + 1) + C



    (3) We can use Long Division:. . \int\frac{1}{1+e^x}\,dx \;=\;\int\frac{1 + e^x - e^x}{1 + e^x}\,dx \;=\;\int\left(\frac{1+e^x}{1+e^x} - \frac{e^x}{1+e^x}\right)\,dx
    . . =\;\int\left(1 - \frac{e^x}{1 + e^x}\right)\,dx \;= \;x - \ln(1 + e^x) + C



    (4) One day, while totally bored, I came up with this substitution . . .

    Let e^x = \tan^2\!\theta \quad\Rightarrow\quad e^{\frac{x}{2}} = \tan\theta \quad\Rightarrow\quad \frac{x}{2} = \ln(\tan\theta) \quad\Rightarrow\quad dx = 2\frac{\sec^2\theta}{\tan\theta}\,d\theta
    Substitute: . \int\frac{2\frac{\sec^2\theta}{\tan\theta}\,d\thet  a}{\sec^2\theta} \;=\;2\int\cot\theta\,d\theta \;=\;2\ln(\sin\theta) + C

    Since \tan\theta = e^{\frac{x}{2}}, then \theta is in a right triangle
    . . with: . opp = e^{\frac{x}{2}} and adj = 1
    Hence: . hyp = \sqrt{1 + e^x} . . . and: . \sin\theta = \frac{e^{\frac{x}{2}}}{\sqrt{1 + e^x}}
    And we have: . 2\ln\left(\frac{e^{\frac{x}{2}}}{\sqrt{1+e^x}}\rig  ht) + C


    And, of course, you are expected to show that these answers are equivalent.

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  9. #9
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    Well if you really want to be ugly I have another way of doing this.

    \frac{1}{1+e^x} = \frac{1}{2+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....}

    Long division.

    Integrate term-by-term

    And bring to standard form of ones of the know functions.

    I leave the details to you.
    Last edited by ThePerfectHacker; June 25th 2007 at 07:52 PM.
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  10. #10
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    Question

    Quote Originally Posted by ThePerfectHacker View Post
    \frac{1}{1+e^x} = \frac{1}{3+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....}
    Where does the "3" come in the denominator?
    It should be "2" I suppose.
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  11. #11
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    Quote Originally Posted by curvature View Post
    Where does the "3" come in the denominator?
    It should be "2" I suppose.
    Yes I fixed it.
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