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Math Help - Approach to this integral

  1. #1
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    Approach to this integral

    This needs to be integrated:

    1/(e^(2x) + e^(x)) w/respect to x

    Can't do partial fractions because e^x never equals zero. Right? Any suggesions? Thanks,

    JN
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  2. #2
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    I would go ahead and write it this way:

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

    Now, it shouldn't be too bad.

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

    The other can be done with a simple u sub:
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  3. #3
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    You may wish to rethink that Partial Fraction idea. I knew there was something wrong with that method of finding Partial Fractions.

    Have you considered \frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} - \frac{1}{e^{x}+1}?
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  4. #4
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    Quote Originally Posted by TKHunny View Post
    You may wish to rethink that Partial Fraction idea. I knew there was something wrong with that method of finding Partial Fractions.

    Have you considered \frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} - \frac{1}{e^{x}+1}?
    Hey thanks for the input. But it seems that I'm having some sort of mathmatical disconnect here:

    1. By factoring: \frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} *\frac{1}{e^{x}+1}
    2. How does this equal \frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} -\frac{1}{e^{x}+1}?
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  5. #5
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    Hello,

    Quote Originally Posted by Jim Newt View Post
    Hey thanks for the input. But it seems that I'm having some sort of mathmatical disconnect here:

    1. By factoring: \frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} *\frac{1}{e^{x}+1}
    2. How does this equal \frac{1}{e^{2x}+e^{x}} = \frac{1}{e^{x}} -\frac{1}{e^{x}+1}?
    \frac{1}{e^{x}(e^{x}+1)}=\frac{1+e^x-e^x}{e^{x}(e^{x}+1)}=\frac{1+e^x}{e^{x}(e^{x}+1)}-\frac{e^x}{e^{x}(e^{x}+1)}=\dots

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  6. #6
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    Ah dang algebra! Thanks!
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jim Newt View Post
    This needs to be integrated:

    1/(e^(2x) + e^(x)) w/respect to x

    Can't do partial fractions because e^x never equals zero. Right? Any suggesions? Thanks,

    JN
    Just remember that if you get stuck you may legally do the following

    Let u=e^x

    So that

    \frac{1}{e^{2x}+e^x}=\frac{1}{e^x(e^x+1)}=\frac{1}  {u(u+1)}

    Now let

    \frac{1}{u(u+1)}=\frac{A}{u}+\frac{B}{u+1}

    Multiplying through by u(u+1) we get

    1=A(u+1)+B(u)

    Now letting u=0

    we get

    1=A(0+1)+B(0)\Rightarrow{1=A}

    and letting u=-1

    we see that

    1=A(1+-1)+B(-1)\Rightarrow{B=-1}


    so we see that

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

    now back subbing we see that

    \frac{1}{e^{2x}+e^x}=\frac{1}{e^x}-\frac{1}{e^x+1}


    and just to finish it

    \int\frac{dx}{e^x}+\int\frac{1}{e^x+1}dx=\int{e^{-x}dx}+\int\frac{e^x+1-e^x}{e^x+1}= -e^{-x}+\int\frac{e^x+1}{e^x+1}dx-\int\frac{e^x}{e^x+1}dx=-e^{-x}+x-\ln|e^x+1|+C
    Last edited by Mathstud28; May 29th 2008 at 08:01 PM.
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