# Thread: trouble with integral by parts

1. ## trouble with integral by parts

I'm trying to solve an integral by parts and got stuck. I'm at step 4 of my work.

I THINK I've got it right but it could be because of how I set up the whole thing that could be throwing me off.

2. Sorry, are you referring to actual integration by parts, cause I don't see the need to use it.

Your original equation can be simplified to $\displaystyle \displaystyle \int \frac{x^3}{(x+19)^2} ~dx$

Long division should give you
$\displaystyle \displaystyle \int -38 ~dx + \displaystyle \int x ~ dx -6859 \displaystyle \int \frac{1}{(x+19)^2} ~dx + 1083 \displaystyle \int \frac{1}{x+19} ~dx$

The first one is integrating a constant. You should be able to do that...

The second one is integrating a single x. That should be easy too

The third one, you can integrate it directly if you're comfortable, or simply use the substitution u = x + 19 (remember the limits change during substitution, so turn the u back into x after integrating or change your limits)

The fourth one integrates into a log.

3. $\displaystyle \displaystyle{\int \frac {x^3}{(x+19)^2}dx}=$

$\displaystyle \displaystyle{\int \frac {( \; (x+19)-19)^3}{(x+19)^2}dx}$

Use this formula

$\displaystyle (a+b)^3=a^3+3a^2b+3ab^2+b^3$

where
$\displaystyle a=x+19 \; \; b=-19$

4. Here's a way to avoid long division:

You have successfully written it as $\displaystyle \displaystyle \int \frac{x^3}{(x+19)^2}\;{dx}$.

Let $\displaystyle y = x+19$, then $\displaystyle x^3 = (y-19)^3$ and $\displaystyle dy = dx$ , so we have:

$\displaystyle \displaystyle \int \frac{x^3}{(x+19)^2}\;{dx} = \int \frac{(y-19)^3}{y^2}\;{dy}$

The rest is routine. Have you tried learning the Latex?
It might make the task of posting and getting help easier.