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Integral question

  1. Aug 10, 2005 #1
    can someone show me how to do this integral:

    [tex] \int \frac{(1-x)}{x^2} e^{x-1} dx[/tex]
     
  2. jcsd
  3. Aug 10, 2005 #2

    TD

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    Integration by parts works this way, but perhaps there's an easier way.

    [tex]\begin{array}{l}
    \int {\frac{{\left( {1 - x} \right)}}{{x^2 }}e^{x - 1} } dx = - \int {\left( {1 - x} \right)e^{x - 1} } d\left( {\frac{1}{x}} \right) = - \left( {\frac{{\left( {1 - x} \right)e^{x - 1} }}{x} - \int {\frac{1}{x}d\left( {\left( {1 - x} \right)e^{x - 1} } \right)} } \right) \\ \\
    = - \frac{{\left( {1 - x} \right)e^{x - 1} }}{x} - \int {\frac{{ - xe^{x - 1} }}{x}dx} = - \frac{{\left( {1 - x} \right)e^{x - 1} }}{x} + e^{x - 1} + C = \frac{{e^{x - 1} }}{x} + C \\
    \end{array}[/tex]
     
  4. Aug 12, 2005 #3

    GCT

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    here's a simpler more "trivial" solution, by the way what's the derivative of e^x/x? hint, hint
    [tex]I= \int \frac{(1-x)}{x^2} e^{x-1} dx[/tex]
    [tex]=e^{-1} \int \frac{e^{x}dx}{x^{2}} -e^{-1} \int \frac{e^{x}dx}{x}[/tex]
    [tex]J=\int \frac{e^{x}dx}{x^{2}} ,~K=\int \frac{e^{x}dx}{x}[/tex]
    using integration by parts
    [tex]K= \frac{e^{x}}{x}+ \int \frac{e^{x}dx}{x^{2}}[/tex]
    [tex]I=e^{-1}J- \frac{e^{-1}e^{x}}{x} -e^{-1}J,~I= \frac{-e^{-1}e^{x}}{x}+C[/tex]
     
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