To analyze f(x) = ln(x² + 2x + 2) – 2tan⁻¹(eˣ), we find its derivative f'(x).

First, simplify the quadratic: x² + 2x + 2 = (x+1)² + 1.

Now, differentiate: f'(x) = (2x+2)/((x+1)²+1) – 2 * (eˣ)/(1+(eˣ)²).

Simplify: f'(x) = 2(x+1)/((x+1)²+1) – 2eˣ/(1+e²ˣ).

Notice that (x+1)²+1 ≥ 1 and 1+e²ˣ ≥ 1. The derivative f'(x) simplifies to 0 for all x, indicating the function is constant. However, this is a special case. Let's check the constant value.

Evaluate f(x) at a point, say x=0: f(0) = ln(0+0+2) – 2tan⁻¹(1) = ln2 – 2*(π/4) = ln2 – π/2.

This is the value of the function for all x. Therefore, f(x) is a constant function, so it is both non-increasing and non-decreasing, but strictly speaking, it is not monotonic in the typical sense of being strictly increasing or decreasing. The least (and only) value is ln2 – π/2.

Final Answer: The least value of f(x) is ln2 – π/2.