# Thread: integration of natural log

1. ## integration of natural log

$\int ln(3x - 7)dx$

2. Originally Posted by wik_chick88
$\int ln(3x - 7)dx$
Set $t= 3x-7 \implies \text{dt}= 3\text{dx}$

$\int \ln(3x - 7)\text{dx} = \frac{1}{3}\int \ln(t)\text{dt}$

$u= \ln(t) \implies du= \frac{1}{t}\text{dt}$
$\text{dv}= \text{dt} \implies v= t$

$\frac{1}{3}\left\{ t \ln(t)- \int t \frac{1}{t} \text{dt} \right\}$

3. Substitute $t=3x-7$, to get :
$\frac{1}{3}\int ln(t) \, dt$
Now, use integration by parts ..

4. the final answer i get is $(x - \frac{7}{3})ln(3x - 7) - \frac{ln(3x - 7)}{3} + C$
but the answer given is $(x - \frac{7}{3})ln(3x - 7) - x + C$

where did i go wrong??

5. How can we tell your mistakes if we did not see your solution .. ?!!

6. Originally Posted by wik_chick88
where did i go wrong??
It would be easier if you post your working so that your mistake can be pointed out, it wasn't helpful with just the final answer posted.

7. Originally Posted by wik_chick88
$(x - \frac{7}{3})ln(3x - 7) - \frac{1}{3}\color{red}{(3x - 7)}$ $+C$

$\color{red}{\text{Here is where you messed up:}}$

$\color{red}{\int t\frac{1}{t} = t = (3x-7)}$
aNon1

8. ok.
$y = \int ln(3x - 7) dx$
let $t = 3x - 7$
$\frac{dt}{dx} = 3,$
so.. $dx = \frac{1}{3} dt$

so...
$y = \frac{1}{3} \int ln t dt$
let $u = ln t$, so $u' = \frac {1}{t}$
let $v' = 1$, so $v = t$

so...
$y = \frac{1}{3} (t ln t - \int \frac{1}{t} dt)$
$= \frac{1}{3} (t ln t - ln t) + C$
$= \frac{t ln t}{3} - \frac{ln t}{3} + C$
$= \frac{(3x - 7) ln (3x - 7)}{3} - \frac{ln(3x - 7)}{3} + C$
$= (x - \frac{7}{3})ln(3x - 7) - \frac{ln (3x - 7)}{3} + C$

9. thankyou thankyou thankyou! silly me!