Integral Question e^(2x) dx / e^(2x) + 3e^x + +2

I have the answer book, don't understand how to get to one step
[int] e^(2x) dx / e^(2x) + 3e^x + 2 let u = e^x, then x = ln u and dx = du/u
[int] u² * du/u / u² + 3u + 2 (one of the U's will cancel on the numerator, leaving udu)
[int] udu / (u+1)(u+2) (the denominator factors)
[int] [-1 / u+1 + 2/u+2]du where is the -1 & +2 numerators coming from?

Thnx for the help, people here have really helped me out. (mech)

Partial faction decomp.

Let u/[(u+1)(u+2)] = A/(u+1) + B/(u+2) ==> u = A(u+2) + B(u + 1).

Setting u = -1 isolates A: -1 = A(1) ==> A = -1

Setting u = -2 isolates B: -2 = B(-1) ==> B = 2.

So u/[(u + 1)(u + 2)] = -1/(u + 1) + 2/(u + 2).