I've tried integration by parts as well as substitution using $u=e^{t}$ but nothing seems to work.
$$\int\left(1-e^{-t}\right) \cdot \exp({e^{t}}) dt$$
Can someone please help? Thanks
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1Hint: substitute $u=e^t-t$ – John Doe Sep 17 '16 at 14:32
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It can be integrated, just as j___d commented. – Claude Leibovici Sep 17 '16 at 14:36
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3@Dr.SonnhardGraubner Sometimes I wonder HOW, and really HOW, you can have so many reputations. Your comments are always stupid, and your answers are always useless. – Enrico M. Sep 17 '16 at 14:40
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i have misreaded the question here was it $$\int(1-e^t)e^{e^t}dt$$ – Dr. Sonnhard Graubner Sep 17 '16 at 14:48
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Factoring, we get: $$I:= \int \left(1-e^{-t}\right)e^{e^t}\,\mathrm d t = \int \left(e^t-1\right)e^{e^t-t}\,\mathrm d t$$ Now substitute $u=e^t-t$ and thus $\dfrac{\mathrm d u}{\mathrm d t}=e^t-1$ to get $$I=\int e^u\,\mathrm d u=e^u+C=\boxed{e^{e^t-t}+C}$$
John Doe
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Hint: $1-e^{-t}=f'(t)+g'(t)f(t)$ with $g(t)=e^t, f(t)=e^{-t}$. see for instance
Dealing with integrals of the form $\int{e^x(f(x)+f'(x))}dx$
Idris Addou
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