Lời giải

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

Vì \( f(0) = 0 \), suy ra \( c = 1 \). 

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

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

\( I = \int_1^3 \sqrt{x^4 + 2x^2}  \, dx = \int_1^3 x\sqrt{x^2 + 2} \, dx \)

\( = \frac{1}{2} \frac{(\sqrt{x^2 + 2})^{\frac{3}{2}}}{\frac{3}{2}}  \Bigg|_1^3 = \frac{1}{3}  (x^2 + 2)\sqrt{x^2 + 2} \Bigg|_1^3 \)

\(  = \frac{1}{3} \Big[ 11\sqrt{11} - 3\sqrt{3} \Big] \)

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{B}} \)