Integral ln(3x-1) dx ?

Question

Used Integration by parts; u=ln(3x-1), du=3/(3x-1) dx, dv=dx, v=x. I used the formula uv - Integral v du. I got past the long division, but I think you have to do it again? I did, and am stuck simplifying.

Answer 1

FWIW I get ((3x - 1)(ln(3x-1) - 1))/3 + C out of it.
Yeah, it's about 4 or 5 sheets of paper ☺

Doug

Answer 2

Integral Of Ln 3x

Answer 3

Here's what I got for the answer:

x*ln(3x-1) - x - (1/3)*ln(3x-1) + C = (x-1/3)*ln(3x-1) - x + C.

I differentiated it, and it got me back to ln(3x-1), so I'm guessing it's the right answer. Everything you say to do sounds right. 3x/3x-1 long-divides into 1 - 1/(3x-1). You do not have to long divide again; the integral of 1 is x and the integral of the second term is (1/3)*ln(3x-1). This is where the terms above come from.

Hope that helps!

Answer 4

Change variable: u = 3x - 1

du = 3dx

(1/3)Integral ln(3x-1) 3dx = (1/3)int ln u du

int ln u = u ln u - o + C

Check this:

(u lun u + u )' = u/u + ln u - 1 = ln u

Just plug 3x - 1 instead of u
Ana

Answer 5

integrate ln3x/2