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Int. by Parts

  1. Apr 17, 2006 #1
    Can I use integration by parts recursively on this?

    [tex]\int (xe^x)(x+1)^{-2}[/tex]
     
  2. jcsd
  3. Apr 17, 2006 #2

    matt grime

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    Have you tried?
     
  4. Apr 17, 2006 #3
    Yeah

    [tex]... = \frac{-xe^x}{x + 1} - \int \frac{e^x}{x + 1} \cdot dx[/tex]

    Right so far?
     
  5. Apr 17, 2006 #4

    arildno

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    Nope.
    [tex]\frac{d}{dx}xe^{x}=e^{x}(x+1)[/tex]
    There is at least one other flaw with your work.
     
  6. Apr 17, 2006 #5
    Oops... don't know what I was thinking there. I got it now, thanks.
     
  7. Apr 17, 2006 #6


    Affirmative.

    Formula for integration by parts by Substitution Rule:
    [tex]\int u dv = uv - \int v du[/tex]
    [tex]u = xe^x \; \; \; dv = \frac{1}{(x + 1)^2} dx[/tex]
    [tex]du = e^x (x + 1) dx \; \; \; v = \left( - \frac{1}{x + 1} \right)[/tex]
    [tex]\int (xe^x) \left[\frac{1}{(x + 1)^2} \right] dx = (xe^x) \left(-\frac{1}{x + 1} \right) - \int \left(-\frac{1}{x + 1} \right)[e^x(x + 1)] dx[/tex]
     
    Last edited: Apr 17, 2006
  8. Apr 18, 2006 #7
    Can this

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

    be further simplified to
    [tex]\int (xe^x) \left[\frac{1}{(x + 1)^2} \right] dx = \frac {e^x}{x+1}[/tex]
     
    Last edited: Apr 18, 2006
  9. Apr 19, 2006 #8

    Affirmative
    [tex]\left(xe^x\right)\left(-\frac{1}{x + 1}\right) - \int \left(-\frac{1}{x + 1}\right)\left[e^x\left(x + 1\right)\right] dx = - \frac{xe^x}{x + 1} + \int e^x dx[/tex]

    [tex]- \frac{xe^x}{x + 1} + \int e^x dx = - \frac{xe^x}{x + 1} + e^x = e^x \left(- \frac{x}{x + 1} + 1\right) = \frac{e^x}{x + 1}[/tex]

    [tex]\boxed{\int \left(xe^x\right)\left[\frac{1}{\left(x + 1\right)^2}\right] dx = \frac{e^x}{x + 1}}[/tex]
     
    Last edited: Apr 19, 2006
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