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Math Help - Definite Intergral

  1. #1
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    Definite Intergral

    Any hints on where to start?

    Integral from 1 to 0 of (x*e^x)/(x+1)^2 dx
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by millerst View Post
    Any hints on where to start?

    Integral from 1 to 0 of (x*e^x)/(x+1)^2 dx
    Make the following observation

    \frac{x\cdot e^x}{(1+x^2)}=\frac{e^x}{1+x}-\frac{e^x}{(1+x)^2}
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  3. #3
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    Or we should say that clearly ,

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


    Let D be   \frac{d}{dx}   ( the differential operator )


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


    can you see something special ?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by simplependulum View Post
    Or we should say that clearly ,

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


    Let D be  \frac{d}{dx} ( the differential operator )


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


    can you see something special ?
    That works too. But I was actually making the observation \frac{xe^x}{(1+x)^2}=\frac{e^x}{1+x}-\frac{e^x}{(1+x^2}=D\left(e^x\right)\cdot\frac{1}{  1+x}+e^x\cdot D\left(\frac{1}{1+x}\right)

    Same thing, two different ways.
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