Evaluate $\int_{-1}^{1} \exp(x+e^{x})\,dx$

Question

Evaluate

$$\int_{-1}^{1} \exp({x+e^{x}})\,dx$$ 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.

Answer 1

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

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