# Thread: Quick question involving integration by parts

1. ## Quick question involving integration by parts

Here is a solution guide to a problem;

$\displaystyle \int \frac {3}{2x-3} + \frac {2}{(2x-3)^5} dx$
$\displaystyle 3*\frac{1}{2} \int \frac {1}{u}du + 2*\frac{1}{2} \int \frac{1}{u^5} du$

The red is what I don't understand. They drew out the numerator because it was a k value, but where did the one half come from?

Has anyone ever really been far even as decided to use even go want to do look more like?

2. Has anyone ever really been far even as decided to use even go want to do look more like?
Whhhaaaaaatttttttttttttt?

3. Originally Posted by maddas
Whhhaaaaaatttttttttttttt?
I was wondering the same thing.

4. Originally Posted by Archduke01
Here is a solution guide to a problem;

$\displaystyle \int \frac {3}{2x-3} + \frac {2}{(2x-3)^5} dx$
$\displaystyle 3*\frac{1}{2} \int \frac {1}{u}du + 2*\frac{1}{2} \int \frac{1}{u^5} du$

The red is what I don't understand. They drew out the numerator because it was a k value, but where did the one half come from?

Has anyone ever really been far even as decided to use even go want to do look more like?
supposing $\displaystyle u = 2x-3$

$\displaystyle du = 2 dx$

$\displaystyle \therefore dx = \frac{du}{2}$

when you replace $\displaystyle 2x-3$ with u, replace $\displaystyle dx$ with $\displaystyle \frac{1}{2} du$. Thats how the $\displaystyle \frac{1}{2}$ comes before the integral

5. Awesome, thanks bumble bro.

PS: Iron Maiden rules.