To find the minimum value of (f(x) = \frac{x^2 + 2x + 4}{x + 1}) where (x > -1), follow these steps:

Step 1: Simplify the function

Let (t = x + 1) (since (x > -1), (t > 0)). Then (x = t - 1). Substitute into the function:
[ f(x) = \frac{(t - 1)^2 + 2(t - 1) + 4}{t} ]
Expand the numerator:
[ (t^2 - 2t + 1) + 2t - 2 + 4 = t^2 + 3 ]
Thus:
[ f(x) = \frac{t^2 + 3}{t} = t + \frac{3}{t} ]

Step 2: Apply AM-GM inequality

For positive (t), the Arithmetic Mean-Geometric Mean inequality gives:
[ t + \frac{3}{t} \geq 2\sqrt{t \cdot \frac{3}{t}} = 2\sqrt{3} ]
Equality holds when (t = \frac{3}{t}), i.e., (t^2 = 3) → (t = \sqrt{3}) (since (t > 0)).

Step 3: Confirm the minimum value

Substitute (t = \sqrt{3}) back to (x): (x = \sqrt{3} - 1) (which is > -1). The minimum value is (2\sqrt{3}).

Answer: (\boxed{2\sqrt{3}})

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