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Basic intergral

  1. Dec 2, 2007 #1
    the intergral of 1/3x either equals 1/3 ln 3x or 1/3 ln x - i dont know which one is correct though because cant you take the 1/3 out of the intergral and then you get 1/3 ln x - so confused - i should really know this...
     
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  3. Dec 2, 2007 #2

    arildno

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    It doesn't matter which one you use, since they only differ by a constant:
    [tex]\frac{1}{3}\ln(3x)=\frac{1}{3}\ln(x)+\frac{1}{3}\ln(3)[/tex]
    where the last term is simply a constant, as claimed.
     
  4. Dec 3, 2007 #3

    HallsofIvy

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    You are wrong about "can't take the 1/3 out of the intergral" ("intergral": Boston accent?). [itex]\int 1/(3x)dx= \int (1/3)(1/x)dx= (1/3)\int (1/x)dx= (1/3) ln x+ c[/itex]

    Of course, you could also let u= 3x so that du= 3 dx and the integral becomes
    [tex]\int (1/(3x)) dx= \int (1/u)(du/3)= (1/3)ln(u)+ C= (1/3)ln(3u)+ C[/tex]
    As arildno points out, since ln(3u)= ln(u)+ ln(3), this just becomes
    (1/3)ln(u)+ (1/3)ln(3)+ C and (1/3)ln(3)+ C is just a different "constant of integration".
     
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