# Quick question involving integration by parts

• Apr 13th 2010, 06:42 PM
Archduke01
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?
• Apr 13th 2010, 06:54 PM
Quote:

Has anyone ever really been far even as decided to use even go want to do look more like?
Whhhaaaaaatttttttttttttt?
• Apr 13th 2010, 06:55 PM
dwsmith
Quote:

Whhhaaaaaatttttttttttttt?

I was wondering the same thing.
• Apr 13th 2010, 07:20 PM
harish21
Quote:

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
• Apr 13th 2010, 07:57 PM
Archduke01
Awesome, thanks bumble bro.

PS: Iron Maiden rules.