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Integration By Parts

  1. Sep 29, 2006 #1
    I'm doing a problem where I'm supposed to use integration by parts. I have:

    Integral ln(x+3)dx

    u=ln(x+3) dv=dx
    du=1/(x+3) v=x

    integral ln(x+3) = xln(x+3) - integral x/(x+3)

    That's as far as I've gotten. I know that I should be able to find the integral of x/(x+3) fairly easily, but I just completely forgot how to do it. Could somebody please lend me a hand?
     
  2. jcsd
  3. Sep 29, 2006 #2
    Do the long division of x / (x + 3). Then you should have an easily integrable result.
     
  4. Sep 29, 2006 #3
    Another option is this: Rewrite your numerator as
    [tex] \int \frac{x + 3 - 3}{x + 3} dx[/tex] . Then, you can have two integrals, [tex] \int \frac{x+3}{x+3} dx -3 \int \frac{1}{x+3} dx [/tex]

    That should be easier to evaluate. Adding the +3 -3 is the same as adding zero, so it's okay.
     
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