[SOLVED] Integration by parts with a natural log

**redman223** · August 31st 2008, 03:22 PM

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

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

I came up with:
$ 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 $\int x~dx$ . I am sure I messed up somewhere.

**Jhevon** · August 31st 2008, 03:41 PM

Originally Posted by redman223

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

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

I came up with:
$ 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 $\int x~dx$ . I am sure I messed up somewhere.

nope, you're correct

now finish up

**skeeter** · August 31st 2008, 03:42 PM

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

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

taking the derivative ...

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

you're back where you started.

**redman223** · August 31st 2008, 06:18 PM

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.

**Jhevon** · August 31st 2008, 06:26 PM

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!

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