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Math Help - Intergrals and e

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    Intergrals and e

    the intergral of: [e^(1/x)]/x^2 dx
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    Quote Originally Posted by calcutron20 View Post
    the intergral of: [e^(1/x)]/x^2 dx
    \frac{e^{\frac{1}{x}}}{x^2} = e^{x^{-1}}x^{-2}.


    To find \int{e^{x^{-1}}x^{-2}\,dx}

    Let u = x^{-1} so that \frac{du}{dx} = -x^{-2}.



    \int{e^{x^{-1}}x^{-2}\,dx} = -\int{e^{x^{-1}}(-x^{-2})\,dx}

     = -\int{e^{u}\,\frac{du}{dx}\,dx}

     = -\int{e^u\,du}

     = -e^{u} + C

     = -e^{x^{-1}} + C

     = -e^{\frac{1}{x}} + C.
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