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Having trouble with a fairly simple integration

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm in the middle of solving a D.E using integration factors, and I've ended up with an integral on the R.H.S which is as follows: [tex] \displaystyle \int \frac {e^ \frac {-1}{x}}{x^4} dx [/tex]


    I've tried to use integration by parts to do it, but I keep having to use it again and again. I've checked that my integrating factor is correct, and it is. I'm just having a hard time doing it. Any help is greatly appreciated. Thanks, Matt.
     
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  3. Feb 21, 2012 #2

    Mark44

    Staff: Mentor

    Try splitting it this way:
    u = x^(-2), dv = x^(-2)e^(-1/x)dx

    I haven't worked it through, so can't guarantee this will work, but this is what I would try.
     
  4. Feb 21, 2012 #3
    Okay, thanks. I'll give it a go.
     
  5. Feb 21, 2012 #4

    dextercioby

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    If you let 1/x = t (new variable), what do you get ?
     
  6. Feb 21, 2012 #5

    Dick

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    I would start by substituting u=(-1/x). The integration by parts should be straightforward. You only need to do it twice. Can you show where you are having problems?
     
  7. Feb 21, 2012 #6
    Using [tex] \displaystyle u= \frac {-1}{x} [/tex] I get [tex] \displaystyle \int e^{u} u^{2} du [/tex]

    Then I used integration by parts to get that equal to [tex] \displaystyle u^{2}e^{u} - \int 2u e^{u} du [/tex]

    Then using integration by parts again I get that equal to [tex] \displaystyle 2u e^{u} - \int 2e^{u} [/tex] which is just [tex] \displaystyle 2u e^{u} - 2e^{u} [/tex]

    Then subbing back in and taking out a factor of [tex] \displaystyle e^u [/tex] I get the answer to the first integral as [tex] \displaystyle e^{u}(u^{2}-2u+2) [/tex]

    And subbing u back in I get [tex] \displaystyle e^{ \frac {-1}{x}}( { \frac {1}{x^{2}}} + { \frac {2}{x}} + 2) [/tex]

    Does that look right?
     
  8. Feb 21, 2012 #7

    Mark44

    Staff: Mentor

    You can check yourself - differentiate what you ended with and you should get the integrand of your integral.
     
  9. Feb 21, 2012 #8
    Yeah I checked it with wolfram, its correct. Thanks for the tip, I'm not that good with integrals so I probably wouldn't have spotted that.
     
  10. Mar 4, 2012 #9

    dextercioby

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    Late I am, but for the record, don't leave out the integration constant, because it shows what you're looking for: the 'family' of antiderivatives.
     
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