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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.

  • 1
    Ah, I see that now. For some reason my mind wasn't in a mode to split up the integrand into two pieces last night.2012-03-13

3 Answers 3

-7

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.

  • 15
    Yes, but it's a terrible way of doing the problem. Not only does it use unnecessarily technology, but the algebra is harder than the elementary solution given above. [moderation edited to remove superfluous insult, M.G.]2012-03-14
10

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

  • 1
    Ah, of course. I glossed over that fact (it's late, that's an excuse, right?) - would much rather keep it simple than double substitutions and partial fractions. Thanks.2012-03-13
8

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} $

  • 0
    @Victor: The quotient rule is for differentiation, not integration. The middle equation here tells us that the fractions can be simplified.2012-03-13