Đáp án:

\(\int_{0}^{1} [f'(x) \cdot (f(x))^2 + 1] dx = 2 \int_{0}^{1} f(x) \cdot \sqrt{f'(x)} dx\)

\(\Leftrightarrow \int_{0}^{1} (\sqrt{f'(x)} \cdot f(x) - 1)^2 dx = 0 \Leftrightarrow \sqrt{f'(x)} \cdot f(x) = 1\)

\(\Leftrightarrow f'(x) \cdot (f(x))^2 = 1\)

\(\Rightarrow \int f'(x) \cdot (f(x))^2 dx = \int dx \Rightarrow \frac{(f(x))^3}{3} = x + C\)

\(\Rightarrow(f(x))^3 = 3x + 3C\)

\(f(0) = 2 \Rightarrow 3C = 8\)

\((f(x))^3 = 3x + 8 \Rightarrow \int_{0}^{1} (f(x))^3 dx = \frac{3x^2}{2} + 8x \big|_{0}^{1} = \frac{19}{2} \Rightarrow \boxed{D}\)