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## Integrating Factor, how do you get this?

As I said, they didn't "drop" mu, they solved for it!
If
$$\frac{du}{dx}= F(x)u$$
then
$$\frac{du}{u}= F(x)dx$$
and, integrating both sides,
$$ln(u)= \int F(x)dx$$

Now take the exponential of both sides.