Từ giả thiết suy ra \(f'\left( {\rm{x}} \right) \ge 0\,,\,\,\forall {\rm{x}} \in \left[ {1\,;\,\,4} \right]\) và \({\rm{f}}\left( {\rm{x}} \right) \ge {\rm{f}}\left( 1 \right) > 0\,,\,\,\forall {\rm{x}} \in \left[ {1\,;\,\,4} \right]\).

Ta có \[x + 2xf\left( {\rm{x}} \right) = {\left[ {f'\left( {\rm{x}} \right)} \right]^2} \Leftrightarrow x\left[ {1 + 2f\left( {\rm{x}} \right)} \right] = {\left[ {f'\left( {\rm{x}} \right)} \right]^2} \Leftrightarrow \frac{{f'\left( {\rm{x}} \right)}}{{\sqrt {1 + 2f\left( {\rm{x}} \right)} }} = \sqrt x \].

Suy ra: \[\int {\frac{{f'\left( {\rm{x}} \right)}}{{\sqrt {1 + 2{\rm{f}}\left( {\rm{x}} \right)} }}} {\rm{dx}} = \int {\sqrt {\rm{x}} dx} \Leftrightarrow \sqrt {1 + 2{\rm{f}}\left( {\rm{x}} \right)} = \frac{2}{3}{\rm{x}}\sqrt {\rm{x}} + {\rm{C}}\].

Vì \(f\left( 1 \right) = \frac{3}{2} \Rightarrow 2 = \frac{2}{3} + C \Leftrightarrow C = \frac{4}{3}.\) Do đó \(f(x) = \frac{1}{2}\left[ {{{\left( {\frac{2}{3}x\sqrt x + \frac{4}{3}} \right)}^2} - 1} \right].\)

Vậy \[I = \int\limits_1^4 {f\left( x \right)dx} = \int\limits_1^4 {\frac{1}{2}\left[ {{{\left( {\frac{2}{3}x\sqrt x + \frac{4}{3}} \right)}^2} - 1} \right]dx} = \frac{{1\,\,186}}{{45}}\]. Chọn D.