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I would like to integrate:

$$ \int{ \frac{dt}{\mathrm{e}^{2\, t}\, \left(\mathrm{e}^{t} + 1\right)} } $$

But I don't know where to start.


The answer according to mupad is

$$ t + \frac{1}{\mathrm{e}^{t}} - \frac{1}{2\, \mathrm{e}^{2\, t}} - \ln\!\left(\mathrm{e}^{t} + 1\right) $$

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up vote 7 down vote accepted

Let $x=e^t$. Then $dx = e^t dt$

Hence, the integral becomes $$\int \frac{dx}{x^3 (x+1)}$$

$$\frac1{x^3(x+1)} = \frac1{x^3} - \frac1{x^2} + \frac1{x} - \frac1{x+1}$$

Now integrate to get the answer

$$\int \frac{dx}{x^3 (x+1)} = \int \frac{dx}{x^3} - \int \frac{dx}{x^2} + \int \frac{dx}{x} - \int \frac{dx}{x+1} = -\frac1{2x^2} + \frac1{x} + \log(x) - \log(x+1)$$

Plug in $x=e^t$ to get

$$\int \frac{dt}{e^{2t} \left(e^t+1\right)} = t + \frac1{e^t} - \frac1{2e^{2t}} - \log(e^t+1)$$

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I might be wrong, but when I did it, I got that the exponential inside of the log was decaying, i.e., it was negative, i.e., $log(e^{-t}+1)$. But please do check to see if thats not what you get. Thanks. P.S. Could you explain where does the first term $t$ come in to play from, and also could you add the constant of integration. I think you missed that piece. – night owl May 31 '11 at 21:55
@nightowl: Both your result and mine are same (the first time is absorbed in your result). $\log(e^{-t}+1) = \log \left( \frac{1+e^t}{e^t} \right) = \log \left( 1+e^t \right) - \log \left( e^t \right) = \log \left(1+e^t \right) - t$ – user17762 May 31 '11 at 22:26
Right, I look at it after my comment and noticed just that. Thanks for the clarification check though. – night owl Jun 1 '11 at 2:08

You can do this just like you would a partial fractions problem. Think of $$\frac{1}{e^{2t}(e^t+1)}$$ as a rational function of the form $$\frac{1}{u^2(u+1)}.$$ Doing partial fractions, we have: $$\begin{align*} \frac{1}{e^{2t}(e^t+1)} &= \frac{A}{e^t} + \frac{B}{e^{2t}} + \frac{C}{e^{t}+1}\\ &= \frac{Ae^t(e^t+1) + B(e^t+1) + Ce^{2t}}{e^{2t}(e^t+1)}\\ &= \frac{(A+C)e^{2t} + (A+B)e^t + B}{e^{2t}(e^t+1)}. \end{align*}$$ So we want $A+C=0$, $A+B=0$, and $B=1$. Therefore, $A=-1$, $C=1$. Therefore, $$\begin{align*} \int\frac{1}{e^{2t}(e^t+1)}\,dt &= \int\left(\frac{-1}{e^t} + \frac{1}{e^{2t}} + \frac{1}{e^t+1}\right)\,dt\\ &= -\int e^{-t}\,dt + \int e^{-2t}\,dt + \int\frac{1}{e^t+1}\,dt. \end{align*}$$ The first two integrals are an easy substitution. The third integral may take a bit more doing, but setting $u=e^t+1$, $du=e^t\,dt$ gives $dt = \frac{1}{u-1}\,du$, so you can do another partial fraction to get $$\int\frac{dt}{e^t+1} = \int\frac{du}{(u-1)u} = \int\left(\frac{1}{u-1} - \frac{1}{u}\right)\,du$$ which is easy to do, yielding the final answer once everything is done.

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$\displaystyle \frac{1}{e^{2t}(e^t+1)} = \frac{(e^t+1)-e^t}{e^{2t}(e^{t}+1)} = \frac{1}{e^{2t}}-\frac{(e^t+1)-e^{t}}{e^t(e^t+1)} = \frac{1}{e^{2t}}-\frac{1}{e^{t}}+\frac{1}{e^{t}+1}.$

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I guess people don't like doing partial fractions without the big method? Hmm... – Lyrebird May 31 '11 at 21:00
Could you share some light on the method you chose not using the "big method" to conclude to the simple answer you got? I.E., the shortcut partial fractions method you choose to go with. – night owl May 31 '11 at 22:02
@night owl - I wrote $1$ as $(e^t+1)-e^t$ twice! That was basically it. – Lyrebird May 31 '11 at 23:43
okay. – night owl Jun 1 '11 at 2:10

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