Evaluating $\int \frac{5-e^{x}}{e^{2x}}dx$ without Partial Fractions

Question

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.

Answer 1

This is (homework), so only hints: $$\begin{eqnarray} \frac{a-b}{c} & = & \frac{a}{c} - \frac{b}{c} \\ \frac{e^{\alpha x}}{e^{\beta x}} & = & ?? \\ \int e^{tx} \ dx & = & ?? \\ \end{eqnarray} $$

Answer 2

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

Answer 3

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.

Answer 4

$$ \begin{aligned} & \int \frac{5-e^{x}}{e^{2 x}} d x \\ =& 5 \int e^{-2 x} d x-\int e^{-x} d x \\ =&-\frac{5 e^{-2 x}}{2}+e^{-x}+C \end{aligned} $$