Why Does Integrating dN/(4-2N) Result in -1/2 ln(|4-2N|)?

[alexis36]

Homework Statement

I have this equation:
integrate: dN/(4-2N)

Homework Equations

The Attempt at a Solution

I am understanding that the dN part just goes away and comes one on the top of the fraction. Then on the bottom, I am still left with 4-2N. I am simply saying that it is equal to ln|4-2N| .. however, I am told that it should be -1/2 ln(|4-2N|) . The -1/2 in front, is that coming from performing a chain rule? And if so, how do you get a chain rule out of this?

 

[rock.freak667]

$$\int \frac{1}{4-2N} dN$$

Let u=4-2N (*) ;$\frac{du}{dN}=-2 \Rightarrow dN=\frac{du}{-2}$$$\int \frac{1}{4-2N} dN \equiv \int \frac{-1}{2} \frac{1}{u} du$$

$$\frac{-1}{2}\int \frac{1}{u} du = \frac{-1}{2}lnu + Constant$$and *