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?
Did you try integrating by parts where $dv = \frac{3}{e^{3x}}$ and $u = 5 - (e^x)$ ? It may be a bit laborious of a calculation,but that's what I'd try if I was instructed NOT to use partial fractions. This is where knowing as many techniques of integration as possible comes in very handy in baby calculus.