$$\int_{-1}^{1} \exp({x+e^{x}})\,dx$$ where $\exp(x)=e^x$. Can anyone give me any tips on where to start with this? I've tried doing it be substitution, with $ u=e^x$ and ended up needing to integrate $ u^{u-1}$, which I don't know how to calculate either.

  • 7
    $\begingroup$ What's the derivative of $e^{e^x}$? $\endgroup$
    – Simon S
    Nov 7 '14 at 16:01

Notive that using following law of exponent

$$a^{b+c}=a^b\cdot a^c$$

Given Integral can be written as $$I=\int e^{x+e^x} \,dx=\int e^{x}e^{e^{x}}\, dx$$ by substituting $e^x=t\iff e^x\,dx=dt$ we get $$I=\int e^{x}e^{e^x} dx=\int e^{t} dt=e^t+C=e^{e^{x}}+C$$

Using above result

$$\int_{-1}^{1} e^{x}e^{e^x} dx=\left[e^{e^{x}}\right]_{-1}^{1}=e^{e}-e^{\frac{1}{e}}$$

  • $\begingroup$ I missed out splitting it into two exponentials before trying substitution. Thanks, I've got it now :) $\endgroup$
    – Megagon
    Nov 7 '14 at 16:20
  • 1
    $\begingroup$ You are fast ;) $\endgroup$ Nov 7 '14 at 17:03

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