integration of natural log

Nov 2009
517
130
Big Red, NY
\(\displaystyle \int ln(3x - 7)dx \)
Set \(\displaystyle t= 3x-7 \implies \text{dt}= 3\text{dx}\)

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

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

\(\displaystyle \frac{1}{3}\left\{ t \ln(t)- \int t \frac{1}{t} \text{dt} \right\}\)
 
Last edited:
Jan 2010
564
242
Kuwait
Substitute \(\displaystyle t=3x-7\), to get :
\(\displaystyle \frac{1}{3}\int ln(t) \, dt\)
Now, use integration by parts ..
 
Apr 2008
204
5
the final answer i get is \(\displaystyle (x - \frac{7}{3})ln(3x - 7) - \frac{ln(3x - 7)}{3} + C\)
but the answer given is \(\displaystyle (x - \frac{7}{3})ln(3x - 7) - x + C\)

where did i go wrong??
 
Jan 2010
564
242
Kuwait
How can we tell your mistakes if we did not see your solution .. ?!!
 
Nov 2009
517
130
Big Red, NY
\(\displaystyle (x - \frac{7}{3})ln(3x - 7) - \frac{1}{3}\color{red}{(3x - 7)}\) \(\displaystyle +C\)

\(\displaystyle \color{red}{\text{Here is where you messed up:}}\)

\(\displaystyle \color{red}{\int t\frac{1}{t} = t = (3x-7)}\)
aNon1
 
Apr 2008
204
5
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...
\(\displaystyle y = \frac{1}{3} (t ln t - \int \frac{1}{t} dt)\)
\(\displaystyle = \frac{1}{3} (t ln t - ln t) + C\)
\(\displaystyle = \frac{t ln t}{3} - \frac{ln t}{3} + C\)
\(\displaystyle = \frac{(3x - 7) ln (3x - 7)}{3} - \frac{ln(3x - 7)}{3} + C\)
\(\displaystyle = (x - \frac{7}{3})ln(3x - 7) - \frac{ln (3x - 7)}{3} + C\)