1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Problematic Integral
  1. A problematic integral (Replies: 16)

  2. Integration of (Replies: 10)

  3. Integral of (Replies: 3)

  4. Integral ? (Replies: 10)

  5. Integral of (Replies: 4)