How do you integrate ln(3x-1)?
Use a u-substitution with u = 3x-1, and the fact that \int ln x dx = x ln x - x + C.
@MITer94 so would it be x ln (3x-1) - x + C or do you have to do something with the chain rule?
No.
@APScholar18 Using u = 3x-1, what is du in terms of dx?
@MITer94 du=3dx
and dx=1/3du
Yes. Substitute x and dx, and integrate in terms of u.
@MITer94 so 3xln(3x-1) - (3x-1) + C or 3xln(3x-1) - (x) + C or x ln (3x-1) - (3x) or are those all wrong?
The answer is (1/3) ((3x-1) ln (3x-1) - (3x-1)) + C. Try actually doing the substitution.
@MITer94 thx that makes sense I kept doing the chain rule for some stupid reason 
ln(3x-1)
u = 3x-1
du/dx = 3, so dx = du/3
ln(3x-1) dx = (1/3)ln(u) du
and integrate
(1/3) * uln(u) -u + C
plug in
(1/3) * (3x-1)ln(|3x-1) - (3x-1) + C
reduce
(x-1/3)ln(|3x-1|) - x + (1/3) + C
@mroadstar thanks.