Just in case you're wondering why the Integrating Factor is [math]\displaystyle e^{P(x)\,dx}[/math], let's look at a general first order linear DE in more detail...
[math]\frac{dy}{dx} + P\,y = Q[/math], where P, Q and y are functions of x.
What we would like is to write the LHS as a single derivative, because then we would be able to solve for y by integrating. To do this, we make use of the Product Rule [math]\displaystyle \frac{d}{dx} \left[ f\,g \right] = f\,\frac{dg}{dx} + g\,\frac{df}{dx}[/math]. At the moment, the LHS doesn't look like a product rule expansion, but it's possible to multiply by a function to make this so. Call this function I. Then
[math]I\,\frac{dy}{dx} + I\,P\,y = I\,Q[/math]
Now in order for the LHS to be a product rule expansion, then that would mean
[math] \begin{align*} \frac{dI}{dx} &= I\,P \\ \frac{1}{I}\,\frac{dI}{dx} &= P \\ \int{\frac{1}{I}\,\frac{dI}{dx}\,dx} &= \int{P\,dx} \\ \int{\frac{1}{I} \,dI } &= \int{P\,dx} \\ \ln{|I|} &= \int{P\,dx} \\ I &= A\,e^{\int{P\,dx}} \end{align*} [/math]
So any value of the constant A will give an integrating factor that works, so we usually just choose the principal value with A = 1. So if we multiply both sides of the DE by [math] \displaystyle e^{\int{P\,dx}}[/math] we find
[math]\displaystyle \begin{align*} e^{\int{P\,dx}}\,\frac{dy}{dx} + e^{\int{P\,dx}}\,P\,y &= e^{\int{P\,dx}}\,Q \\ \frac{d}{dx} \left( e^{\int{P\,dx}}\,y \right) &= e^{\int{P\,dx}}\,Q \\ e^{\int{P\,dx}}\,y &= \int{e^{\int{P\,dx}}\,Q\,dx} \\ y &= e^{-\int{P\,dx}}\int{e^{\int{P\,dx}}\,Q\,dx} \end{align*}[/math]
Hope that helps...