# Integration By Parts

1. Sep 29, 2006

### Frillth

I'm doing a problem where I'm supposed to use integration by parts. I have:

Integral ln(x+3)dx

u=ln(x+3) dv=dx
du=1/(x+3) v=x

integral ln(x+3) = xln(x+3) - integral x/(x+3)

That's as far as I've gotten. I know that I should be able to find the integral of x/(x+3) fairly easily, but I just completely forgot how to do it. Could somebody please lend me a hand?

2. Sep 29, 2006

### steelphantom

Do the long division of x / (x + 3). Then you should have an easily integrable result.

3. Sep 29, 2006

### EbolaPox

Another option is this: Rewrite your numerator as
$$\int \frac{x + 3 - 3}{x + 3} dx$$ . Then, you can have two integrals, $$\int \frac{x+3}{x+3} dx -3 \int \frac{1}{x+3} dx$$

That should be easier to evaluate. Adding the +3 -3 is the same as adding zero, so it's okay.