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

  • 1
    $\begingroup$ Hint: substitute $u=e^t-t$ $\endgroup$
    – John Doe
    Sep 17, 2016 at 14:32
  • $\begingroup$ It can be integrated, just as j___d commented. $\endgroup$ Sep 17, 2016 at 14:36
  • 3
    $\begingroup$ @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. $\endgroup$
    – Enrico M.
    Sep 17, 2016 at 14:40
  • $\begingroup$ i have misreaded the question here was it $$\int(1-e^t)e^{e^t}dt$$ $\endgroup$ Sep 17, 2016 at 14:48

2 Answers 2


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 $$I=\int e^u\,\mathrm d u=e^u+C=\boxed{e^{e^t-t}+C}$$

  • $\begingroup$ Bingo ! You have good eyes. $\endgroup$ Sep 17, 2016 at 14:53
  • $\begingroup$ Thank you for solving it😄 $\endgroup$
    – user359618
    Sep 17, 2016 at 14:54
  • $\begingroup$ Thanks @Claude and you're welcome OP! $\endgroup$
    – John Doe
    Sep 17, 2016 at 15:00

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 $\int{e^x(f(x)+f'(x))}dx$


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