# Thread: [SOLVED] Integration by parts with a natural log

1. ## [SOLVED] Integration by parts with a natural log

$\displaystyle \int 2xln(3x)~dx$

I set:
u=ln(3x)
dv=2x dx
v=x^2
du=1/x

I came up with:
$\displaystyle x^2ln(3x) - \int x^2 * 1/x~dx$

But I can't figure out what to do with the part under the integral, because after multiplying I am left with $\displaystyle \int x~dx$. I am sure I messed up somewhere.

2. Originally Posted by redman223
$\displaystyle \int 2xln(3x)~dx$

I set:
u=ln(3x)
dv=2x dx
v=x^2
du=1/x

I came up with:
$\displaystyle x^2ln(3x) - \int x^2 * 1/x~dx$

But I can't figure out what to do with the part under the integral, because after multiplying I am left with $\displaystyle \int x~dx$. I am sure I messed up somewhere.
nope, you're correct

now finish up

3. you are ok ... $\displaystyle \int x \, dx = \frac{x^2}{2} + C$

you finally get $\displaystyle x^2\ln(3x) - \frac{x^2}{2} + C$

taking the derivative ...

$\displaystyle x^2 \cdot \frac{1}{x} + 2x \cdot \ln(3x) - x = 2x \cdot \ln(3x)$

you're back where you started.

4. Wow, I can't believe that I didn't see that earlier, thanks. I guess my brain is taking a little while longer to come off of summer break. It's amazing how much can be forgotten in as little as 6 weeks.

5. Originally Posted by redman223
Wow, I can't believe that I didn't see that earlier, thanks. I guess my brain is taking a little while longer to come off of summer break. It's amazing how much can be forgotten in as little as 6 weeks.
then you know how bad i have it! my summer break was more than twice that long!