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?


  • 2
    $\begingroup$ Your substitution would work, but the approaches discussed in the answers are simpler. For example, let $u=e^x$ as you did. Then your integral becomes $\int \frac{(5-u)}{u^3}\,du$. This is $\int \left(\frac{5}{u^3}-\frac{1}{u^2}\right)du$, which is very doable without partial fractions. $\endgroup$ – André Nicolas Mar 13 '12 at 4:39
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    $\begingroup$ 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. $\endgroup$ – Joe Mar 13 '12 at 21:02

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

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    $\begingroup$ Ah, wow. I was simply overthinking the problem! Thanks J.D. $\endgroup$ – Joe Mar 13 '12 at 4:09
  • $\begingroup$ Do you have to use the quotient rule in order to finish with this problem? $\endgroup$ – Victor Mar 13 '12 at 6:10
  • $\begingroup$ @Victor: The quotient rule is for differentiation, not integration. The middle equation here tells us that the fractions can be simplified. $\endgroup$ – anon Mar 13 '12 at 9:30

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

  • 1
    $\begingroup$ 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. $\endgroup$ – Joe Mar 13 '12 at 4:10

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.

  • 3
    $\begingroup$ No, I did not try integrating by parts - my class is just learning antiderivatives of functions involving e, so I figured that method was also implied not to be used, as the class has not learned it. Nonetheless, I see how that would also work. Rewriting the integrand and splitting it up into two integrals seems to be the easiest way from what I could tell though. $\endgroup$ – Joe Mar 13 '12 at 21:00
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    $\begingroup$ You would really use integration by parts rather than the suggestion in the earlier answers of distributing the division and integrating simple exponentials? $\endgroup$ – Jonas Meyer Mar 14 '12 at 3:41
  • $\begingroup$ @Jonas Uh,yeah,if I feel more comfortable doing it that way,why not?Either way arrives at the same result. $\endgroup$ – Mathemagician1234 Mar 14 '12 at 4:06
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    $\begingroup$ 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.] $\endgroup$ – Adam Smith Mar 14 '12 at 6:38

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