Đáp án

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

\(\Rightarrow \int \frac{f'(x) f(x)}{\sqrt{1 + (f(x))^2}} dx = x^2 + C \quad \left(\frac{1}{2} \int \frac{u'}{\sqrt{u}} dx = \frac{1}{2} \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} = \sqrt{u}\right)\)  

\(\Rightarrow \sqrt{1 + (f(x))^2} = x^2 + C.\)  

\(f(0) = 0 \Rightarrow C = 1\)  

\(1 + (f(x))^2 = (x^2 + 1)^2 = x^4 + 2x^2 + 1\)  
\(\Rightarrow (f(x))^2 = x^4 + 2x^2 \Rightarrow f(x) = \sqrt{x^4 + 2x^2}\)  

\(\text{Max } f(x) \text{ trên } [1, 3] = f(3) = \sqrt{81 + 18} = \sqrt{99} = 3\sqrt{11} \Rightarrow \boxed{D}\)  

\( f(x) > 0, \forall x \in [1, 3] \rightarrow \text{Max } f(x)_{[1, 3]} = f(3) = 3\sqrt{11}\)

\(f'(x) = \frac{4x^3 + 4x}{2\sqrt{x^4 + 2x^2}} = \frac{2x(x^2 + 1)}{\sqrt{x^4 + 2x^2}} = 0 \Leftrightarrow x = 0\)