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Math Help - having trouble with integral

  1. #1
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    having trouble with integral

    \int{\frac{e^\frac{1}{x}}{x^3}dx}

    I initially tried solving with u substitution, u = 1/x and du = -1/x^2 dx.

    But after I substitute, I'm not sure how to solve:

    \int{\frac{e^u}{u}du}
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  2. #2
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    Quote Originally Posted by absvalue View Post
    \int{\frac{e^\frac{1}{x}}{x^3}dx}

    I initially tried solving with u substitution, u = 1/x and du = -1/x^2 dx.

    But after I substitute, I'm not sure how to solve:

    \int{\frac{e^u}{u}du}
    Lets do this as a sequence of two substitutions first let

    \displaystyle u=\frac{1}{x^2} \implies du=-\frac{2}{x^3}dx

    notice that  \displaystyle \sqrt{u}=\frac{1}{x} this gives

    \displaystyle -\frac{1}{2}\int e^{\sqrt{u}}du

    Now let t=\sqrt{u} \implies dt=\frac{1}{2}\cdot \frac{1}{\sqrt{u}}du=\frac{1}{2}\frac{1}{t}du \iff du=2tdt

    Putting this in gives

    \displaystyle -\frac{1}{2}\int e^{\sqrt{u}}du=-\int t e^{t}dt

    Now just integrate by parts and back substitute

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