# Quick question involving integration by parts

• Apr 13th 2010, 07:42 PM
Archduke01
Quick question involving integration by parts
Here is a solution guide to a problem;

$\int \frac {3}{2x-3} + \frac {2}{(2x-3)^5} dx$
$3*\frac{1}{2} \int \frac {1}{u}du + 2*\frac{1}{2} \int \frac{1}{u^5} du" alt="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, 07:54 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, 07:55 PM
dwsmith
Quote:

Whhhaaaaaatttttttttttttt?

I was wondering the same thing.
• Apr 13th 2010, 08:20 PM
harish21
Quote:

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

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

$du = 2 dx$

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

when you replace $2x-3$ with u, replace $dx$ with $\frac{1}{2} du$. Thats how the $\frac{1}{2}$ comes before the integral
• Apr 13th 2010, 08:57 PM
Archduke01
Awesome, thanks bumble bro.

PS: Iron Maiden rules.