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Integration by parts help natural log hurry tired

  1. Sep 9, 2008 #1
    Integrate the function. (3x)/(3x-2)

    I am spose to use integration by parts but i don't know how to integrate 1/(3x-2). anybody help!?!?
  2. jcsd
  3. Sep 9, 2008 #2
    [tex]\int \frac{dx}{3x-2}[/tex] well, i dont know why ure supposed to use integ. by parts, since a nice subst. would work.

    substitute 3x-2=u, then 3dx=du=> dx=du/3 so

    [tex]\int \frac{dx}{3x-2}=\frac{1}{3}\int \frac{du}{u}=\frac{1}{3}ln|u|+c[/tex] now just go back to the original variable x.
  4. Sep 9, 2008 #3
    Well, i assume now that you were asked to use integ. by parts on the original function. but if you are not required to do so, there is a nice trick to avoid it ,so you will only need to use subst.
  5. Sep 9, 2008 #4
    I tried but it said the answer doesn't involve absolute value... :( the original problem was find the integral of >>> ln(3x-2)
  6. Sep 9, 2008 #5
    well, if you drop the absolute values, then you are assuming that 3x-2>0, otherwise that would not hold.
  7. Sep 9, 2008 #6
    But wait i asked to integrate (3x)/(3x-2) not dx/(3x-2)... hm... I am confused
  8. Sep 10, 2008 #7


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    Homework Helper

    You should have told us so explicitly. Use integration by parts here, denote u=ln(3x-2). Find v and du, and then you'll have to integrate 3x/(3x-2) which you should do by first simplifying the expression by polynomial long division. And are you sure the answer doesn't have absolute signs? There is a ln in the answer is there not?
    Last edited: Sep 10, 2008
  9. Sep 10, 2008 #8
    Yeah, but look here

    [tex] \frac{3x}{3x-2}=\frac{3x-2+2}{3x-2}=\frac{3x-2}{3x-2}+\frac{2}{3x-2}=1+\frac{2}{3x-2}[/tex]

    so all you need to integrate is what i already told you how!...lol....
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