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.