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Problematic Integral

  1. Jan 13, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \left( \frac{1}{x} - \frac{1}{x^2} \right)e^x ~dx[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I tried integrating by parts,
    [tex] \]\int \left( \frac{1}{x} - \frac{1}{x^2} \right)e^x ~dx\\
    Let ~\frac{dv}{dx}=\left( \frac{1}{x} - \frac{1}{x^2} \right), and ~u=e^x.\\
    \therefore v=\ln{x} + \frac{1}{x}, and ~\frac{du}{dx}=e^x\\
    \therefore \int \left( \frac{1}{x} - \frac{1}{x^2} \right)e^x ~dx
    = e^x(\ln{x} + \frac{1}{x}) - \int{ e^x(\ln{x} + \frac{1}{x})}~dx\[ [/tex]

    But I can't see what to do now; the next integral is even messier than the first!
  2. jcsd
  3. Jan 13, 2009 #2


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    Gold Member

    I think that it won't work here, you need to develop e^x power series expansion.
  4. Jan 13, 2009 #3
    First break the integral apart thus:

    [tex]\int \left( \frac{1}{x} - \frac{1}{x^2} \right) e^x\ \text{d}x = \int \frac{1}{x} e^x \ \text{d}x - \int \frac{1}{x^2} e^x \ \text{d}x[/tex]

    Now, integrate the first of the two resulting integrals by parts, letting [itex]u=\frac{1}{x}[/itex] and [itex]\text{d}v=e^x\ \text{d}x[/itex]. You should get a very convenient cancellation.
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