Let $$f(x) = 2x + \tan^{-1}(x)$$ and $$g(x) = \log_e(\sqrt{1+x^2} + x), \quad x \in [0, 3]$$. Then

$$f(x) = 2x + \tan^{-1}(x), \quad g(x) = \log_e(\sqrt{1+x^2} + x), \quad x \in [0, 3]$$

Differentiating both functions:

$$f'(x) = 2 + \frac{1}{1+x^2}$$

$$g'(x) = \frac{1}{\sqrt{1+x^2}+x} \cdot \left(\frac{2x}{2\sqrt{1+x^2}} + 1\right) = \frac{1}{\sqrt{1+x^2}}$$

Comparing derivatives for $$x \in [0, 3]$$:

$$\text{Since } \frac{1}{1+x^2} > 0 \implies f'(x) > 2$$

$$\text{Since } \sqrt{1+x^2} \ge 1 \implies 0 < g'(x) \le 1$$

$$f'(x) > g'(x) \quad \forall x \in [0,3]$$

Analyzing function values from derivatives:

$$\text{Let } h(x) = f(x) - g(x) \implies h'(x) = f'(x) - g'(x) > 0$$

$$h(x) \text{ is strictly increasing}$$

$$h(0) = f(0) - g(0) = 0 \implies h(x) > 0 \quad \forall x \in (0, 3]$$

$$f(x) > g(x) \quad \forall x \in (0,3]$$

Evaluating maximum values since both functions are strictly increasing:

$$\max f(x) = f(3) = 6 + \tan^{-1}(3)$$

$$\max g(x) = g(3) = \log_e(\sqrt{10} + 3)$$

$$\max f(x) > \max g(x)$$