Integration by parts help natural log hurry tired

[darthxepher]

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??

 

[sutupidmath]

\int \frac{dx}{3x-2} well, i don't 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

\int \frac{dx}{3x-2}=\frac{1}{3}\int \frac{du}{u}=\frac{1}{3}ln|u|+c now just go back to the original variable x.

 

[sutupidmath]

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.

 

[darthxepher]

I tried but it said the answer doesn't involve absolute value... :( the original problem was find the integral of >>> ln(3x-2)

 

[sutupidmath]

well, if you drop the absolute values, then you are assuming that 3x-2>0, otherwise that would not hold.

 

[darthxepher]

But wait i asked to integrate (3x)/(3x-2) not dx/(3x-2)... hm... I am confused

 

[Defennder]

I tried but it said the answer doesn't involve absolute value... :( the original problem was find the integral of >>> ln(3x-2)

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?

 

[sutupidmath]

But wait i asked to integrate (3x)/(3x-2) not dx/(3x-2)... hm... I am confused

Yeah, but look here

\frac{3x}{3x-2}=\frac{3x-2+2}{3x-2}=\frac{3x-2}{3x-2}+\frac{2}{3x-2}=1+\frac{2}{3x-2}

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