How to evaluate the integral $\int_0^{\ln3} e^{x-e^x}\,\mathrm dx$?

Question

How to evaluate the following definite integral? $$\int_0^{\ln3} e^{x-e^x}\,\mathrm dx.$$

Should I use some sort of U Substitution?

Answer 1

Use the substitution $u=\mathrm{e}^{x}$, and $\mathrm{d}u=\mathrm{e}^x$, then we get \begin{align*} \int_0^{\ln3}\mathrm{e}^{x-\mathrm{e}^x}\,\mathrm{d}x &= \int_1^3\mathrm{e}^{-u}\,\mathrm{d}u \\ &=\left[-\mathrm{e}^{-u}\right]_1^3 \\ &=\frac{\mathrm{e}^2-1}{\mathrm{e}^3} \end{align*}