Question

Please help me get started with the following integral (click to see).

Answer 1

\[\int\limits_{}^{}\ln(3x-2)dx\]

Answer 2

i chose...
\[u=\ln(3x-2)\]
and
\[v=\int\limits\limits_{}^{}1dx=x\]

Answer 3

or is there a better way to solve this problem...i'm getting stuck on the second time through integrating by parts

Answer 4

this is where i am stuck...
\[=xln(3x-2)-3\int\limits_{}^{}x \frac{ 1 }{ 3x-2 }dx\]

Answer 5

ok, u just stuck in part -3 int (x/(3x-2)) dx, right ?

Answer 6

yea

Answer 7

do integral bu u-sub again
u = 3x-2 ---> x=(u+2)/3
du = 3 dx
dx = du/3
-3 int(x/(3x-2) dx = -3 int(u+2)/3u du = - int (u+2)/u du

Answer 8

* by

Answer 9

then for all elements of numerator divided by u
= - int (1 + 2/u) du
= - int 1 du - int 2/u du
= - u - 2ln(u) + c
subs back that u=(3x-2)
= -(3x-2) -2ln(3x-2) + c

Answer 10

if u=3x-2 and dx=du/3 then...
\[xln(3x-2)-\int\limits_{}^{}\frac{ x }{ u }du\]