Evaluating $\int \frac{5-e^{x}}{e^{2x}}dx$ without Partial Fractions I am trying to evaluate

$$\int \frac{5-e^{x}}{e^{2x}} \mathrm dx$$

I tried rewriting the integral by throwing $e^{2x}$ up on top and using $$u=e^{x}$$ $$du = e^{x} dx$$
I then tried another substitution where $v = 5-u$ 
and $dv = -1 du$ but then I can only simplify the integral to 
$$\int \frac{v}{(v-5)^{3}} \mathrm dv$$
Which would then require partial fractions, which my class has not gotten to quite yet (so I'm not allowed to use the method for homework, sadly).
Is there a simple substitution I am overlooking from the beginning or something?
Thanks.

Rewrite the integrand $5e^{-2x}+e^{-x}$. You know how to integrate $ae^{bx}$ right?
