Concept: Area between curve \(y=\ln(x+e)\), x-axis, and y-axis. Find x-intercept, set up integral, use substitution.
Step 1: x-intercept: \(\ln(x+e)=0 \Rightarrow x+e=1 \Rightarrow x=1-e\). Region from \(x=1-e\) to \(x=0\).
Step 2: Area = \(\int_{1-e}^0 \ln(x+e)dx\). Let \(u=x+e\), \(du=dx\). Limits: \(x=1-e \Rightarrow u=1\); \(x=0 \Rightarrow u=e\). Integral = \(\int_1^e \ln u\,du = [u\ln u - u]_1^e = (e\cdot1 - e) - (1\cdot0 - 1) = 0 - (-1) = 1\).
Step 3: Write the final answer. \(\boxed{1}\)