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

$$f(x) = 2x + \tan^{-1} x$$

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

$$f'(x) \ge 2 + 0 > 0, \quad \forall x \in [0, 3]$$

This means $$f(x)$$ is a strictly increasing function on $$[0, 3]$$

$$\max f(x) = f(3) = 2(3) + \tan^{-1}(3) = 6 + \tan^{-1}(3)$$

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

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

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

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

For $$x \in [0, 3]$$, $$\sqrt{1+x^2} \ge 1$$, which means $$g'(x) > 0$$. Therefore, $$g(x)$$ is also a strictly increasing function on $$[0, 3]$$.

$$\max g(x) = g(3) = \log_e\left(\sqrt{1 + 3^2} + 3\right) = \log_e\left(\sqrt{10} + 3\right)$$

$$6 + \tan^{-1}(3) > \log_e\left(\sqrt{10} + 3\right)$$

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