$\displaystyle \int ln(3x - 7)dx $
Set $\displaystyle t= 3x-7 \implies \text{dt}= 3\text{dx}$
$\displaystyle \int \ln(3x - 7)\text{dx} = \frac{1}{3}\int \ln(t)\text{dt}$
$\displaystyle u= \ln(t) \implies du= \frac{1}{t}\text{dt}$
$\displaystyle \text{dv}= \text{dt} \implies v= t$
$\displaystyle \frac{1}{3}\left\{ t \ln(t)- \int t \frac{1}{t} \text{dt} \right\}$
ok.
$\displaystyle y = \int ln(3x - 7) dx$
let $\displaystyle t = 3x - 7$
$\displaystyle \frac{dt}{dx} = 3,$
so.. $\displaystyle dx = \frac{1}{3} dt$
so...
$\displaystyle y = \frac{1}{3} \int ln t dt$
let $\displaystyle u = ln t$, so $\displaystyle u' = \frac {1}{t}$
let $\displaystyle v' = 1 $, so $\displaystyle v = t$
so...
$\displaystyle y = \frac{1}{3} (t ln t - \int \frac{1}{t} dt)$
$\displaystyle = \frac{1}{3} (t ln t - ln t) + C$
$\displaystyle = \frac{t ln t}{3} - \frac{ln t}{3} + C$
$\displaystyle = \frac{(3x - 7) ln (3x - 7)}{3} - \frac{ln(3x - 7)}{3} + C$
$\displaystyle = (x - \frac{7}{3})ln(3x - 7) - \frac{ln (3x - 7)}{3} + C$