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I have to solve the integral $$\int e^{3t-e^t} dt$$

I tried to substitute $y=3t-e^t$ in the integral by adding $\frac{3-e^t}{3-e^t}$ but in this way I obtain:

$$\int e^{y} \frac{1}{3-e^t} dy$$ I don't know how to get rid of the denominator. I most probably did the wrong substitution. Can someone help me please?

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We have $$\int e^{3t-e^t} dt,$$ let $y = e^t$, therefore $dy = e^tdt = ydt$ and hence $$\int e^{3t-e^t} dt = \int e^{3\ln y - y}\frac{1}{y}dy = \int e^{3\ln y}e^{- y}\frac{1}{y}dy=\int y^2e^{- y}dy,$$ can you go from here?

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  • $\begingroup$ yes thank you!! $\endgroup$ – Ergo Feb 23 '16 at 16:41
  • $\begingroup$ Always welcome! $\endgroup$ – echzhen Feb 23 '16 at 16:42
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$$\int dt \, e^{3 t-e^{t}} = \int d(e^{t}) \, (e^{t})^2 e^{-e^{t}} = \int dy \, y^2 e^{-y}$$

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