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Math Help - integration with e^x

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    integration with e^x

    integrate 1+ e^x/(1-e^x)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by twilightstr View Post
    integrate 1+ e^x/(1-e^x)
    do you mean \int \left( 1 + \frac {e^x}{1 - e^x} \right) ~dx or \int \frac {1 + e^x}{1 - e^x}~dx ?

    in either case, consider the substitution u = 1 - e^x. (for the second integral, it may be easier to use this substitution if you multiply by \frac {e^x}{e^x} first)
    Last edited by Jhevon; February 8th 2009 at 01:40 AM.
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    do you mean \int \left( 1 + \frac {e^x}{1 - e^x} \right) ~dx or \int \frac {1 + e^x}{1 - e^x}~dx ?
    Either way, make the substitution u = e^x.
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    i dont know how to integrate (1+u)/(u-u^2)
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    Like a stone-audioslave ADARSH's Avatar
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    Watch this
     <br />
u= e^{x}<br />

     <br />
\frac{1+e^x} {1-e^x} = \frac{1+u}{1-u}<br /> <br />

    \int{\frac{1+u}{u(1-u)}du}

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

    \int{\frac{u+1-u}{u(1-u)}du} +\int{\frac{1}{(1-u)}du}


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

    \int{\frac{1}{(1-u)}du} + \int{\frac{1}{u}du} +\int{\frac{1}{(1-u)}du}
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