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Improper integral

  1. Feb 21, 2008 #1
    I used u substitution to get to this point:

    [tex]\lim_{R\rightarrow\ 3-}} \int_{-3}^{R-3} (\frac{u + 3}{sqrt{u}}) du[/tex]

    is the only way to proceed from here using integration by parts?
     
    Last edited: Feb 21, 2008
  2. jcsd
  3. Feb 21, 2008 #2
    If your calculations are correct up to this point, i do not think you need to use integ by parts. Simply just try to rearrange the integrand like this

    (u+3)/u^1/2= u/u^1/2 +3/u^1/2 = u^1-1/2 +3u^-1/2 and i guess you will be fine!!!!
     
  4. Feb 21, 2008 #3
    ahhh, I didn't think to break up the fraction, for some reason I missed that

    thanks :)
     
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