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Thread: A boring limit that I'm struggling with

  1. #1
    Super Member angel.white's Avatar
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    A boring limit that I'm struggling with

    I know the answer, but I'm not getting at it, can someone point out my error? It's probably something stupid, b/c I've been doing math all day and my brain is fried.


    Correct answer should be $\displaystyle x-ln(e^x+1)+C$

    I'm not so interested in the best way to get this answer as I am in what I did incorrect. I did it 3 times now, and got the same answer (granted I used the same method, but this implies I also made the same mistake)
    ================================================

    $\displaystyle \int \frac 1{1+e^x} ~dx$
    --------------
    Substitute
    $\displaystyle a = 1+e^x$

    $\displaystyle =\frac 1{a-1} ~da = dx$
    --------------

    $\displaystyle =\int \frac 1{a-1}*\frac 1a ~da$

    $\displaystyle =\int \frac 1{(a-1/2)^2+1/4}\ ~da$
    --------------
    Substitute
    $\displaystyle \frac 12tan(b) = a-\frac 12$

    $\displaystyle \frac 12sec^2b ~db = da$
    --------------


    $\displaystyle =\frac 12\int sec^2b*\frac 1{\frac 14(tan^2b+1)}~db$

    $\displaystyle =2 \int~db$

    $\displaystyle =2b+c$

    $\displaystyle =2 ~arctan(2a-1)+c$

    $\displaystyle =2 ~arctan(2e^x+1)+c$
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  2. #2
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    $\displaystyle \frac{1}{e^x+1} = \frac{1}{e^x(e^x+1)} \cdot (e^x)'$.
    Let $\displaystyle t = e^x$ and so we get $\displaystyle \int \frac{1}{t(t+1)} dt$.
    Use partial fractions.
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Use partial fractions.
    It seems so obvious when you say it >.<


    I'll use this method to get the correct answer, but can you tell me where I went wrong in my methods? presumably there is a discrepancy in my work (aside from using a less effective technique), and I want to see where it is, because I have made the same error at least 3 times while working on this (more if you count all the times I checked my work).
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  4. #4
    o_O
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    $\displaystyle \frac{1}{a-1} \cdot \frac{1}{a}$

    $\displaystyle =\frac{1}{a^{2} - a}$

    $\displaystyle =\frac{1}{\left(a^{2} - a {\color{blue} + \frac{1}{4}}\right) {\color{blue} - \frac{1}{4}}}$

    $\displaystyle = \frac{1}{\left(a - \frac{1}{2}\right)^{2} {\color{blue} - } \frac{1}{4}}$

    That might've thrown a wrench to your sub? Seems messy from here
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  5. #5
    Super Member angel.white's Avatar
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    oh, thank you, it was driving me nuts not seeing my mistake.
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  6. #6
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    Quote Originally Posted by angel.white View Post
    It seems so obvious when you say it >.<
    Ever watched pairs skating in the Winter Olympics? They make it look easy. Same idea here.
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  7. #7
    o_O
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    I know TPH is godly when it comes to integration around these parts but didn't he just do a simple substitution of $\displaystyle u = 1 + e^{x}$
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  8. #8
    Super Member angel.white's Avatar
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    Quote Originally Posted by o_O View Post
    I know TPH is godly when it comes to integration around these parts but didn't he just do a simple substitution of $\displaystyle u = 1 + e^{x}$
    Well he also pointed out that a partial fraction decomposition would be much simpler than a trig substitution like I was doing (though I used the wrong substitution, should have used secant, that was the error that stemmed from writing that negative 1/4 as a positive 1/4 >.< )
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  9. #9
    o_O
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    That's what I mean:
    $\displaystyle u = 1+ e^{x}$
    $\displaystyle du = e^{x}dx = (u-1) du$

    So with the subs:
    $\displaystyle \int \frac{1}{u (u - 1)}du$

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  10. #10
    Super Member angel.white's Avatar
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    I will try it again using partial fractions, it's a new day, I think we can get it to work ^_^


    $\displaystyle \int \frac 1{1+e^x} ~dx$

    ---------------
    Substitute:
    $\displaystyle a = 1+e^x$

    $\displaystyle \frac 1{a-1}da = dx$
    ---------------

    $\displaystyle =\int \frac 1{a(a-1)} ~da$

    Partial Fraction Decomposition

    $\displaystyle =\int \left(\frac 1{a-1} -frac 1a \right)~da$

    $\displaystyle =ln|a-1| -ln|a| +C$

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

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

    Hooray, it worked !
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  11. #11
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    $\displaystyle \int {\frac{1}
    {{1 + e^x }}\,dx} = \int {dx} - \int {\frac{{e^x }}
    {{1 + e^x }}\,dx} = x - \ln \left( {1 + e^x } \right) + k.$
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  12. #12
    Super Member angel.white's Avatar
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    Quote Originally Posted by Krizalid View Post
    $\displaystyle \int {\frac{1}
    {{1 + e^x }}\,dx} = \int {dx} - \int {\frac{{e^x }}
    {{1 + e^x }}\,dx} = x - \ln \left( {1 + e^x } \right) + k.$
    I always find your approaches to be so inspiring, thank you ^_^
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