Answer
$$\frac{1}{e}-\frac{1}{e^3}.$$
Work Step by Step
Let $ u=e^x $, then $ du=e^xdx $, $ x:=0\to \ln 3$, $ u:=1\to 3$.
Now, we have
$$\int_0^{\ln 3} e^{x-e^x}dx =\int_0^{\ln 3} e^{x}e^{-e^x}dx=\int_1^3 e^{-u}du\\
= -e^{-u}|_1^3=-e^{-3}+e^{-1}=\frac{1}{e}-\frac{1}{e^3}.$$
