Lời giải chi tiết:

Ta có:

\(\begin{array}{l}x + 2xf\left( x \right) = {\left[ {f'\left( x \right)} \right]^2}\,\,\,\forall x \in \left[ {1;4} \right]\\ \Leftrightarrow f'\left( x \right) = \sqrt {x\left[ {1 + 2f\left( x \right)} \right]} \,\,\,\forall x \in \left[ {1;4} \right]\\ \Leftrightarrow f'\left( x \right) = \sqrt x .\sqrt {1 + 2f\left( x \right)} \,\,\,\forall x \in \left[ {1;4} \right]\\ \Leftrightarrow \sqrt x  = \dfrac{{f'\left( x \right)}}{{\sqrt {1 + 2f\left( x \right)} \,}}\\ \Leftrightarrow \sqrt x  = \dfrac{{\left[ {1 + 2f\left( x \right)} \right]'}}{{2\sqrt {1 + 2f\left( x \right)} \,}}\end{array}\)

Lấy nguyên hàm hai vế ta được: \(\int {\sqrt x dx}  = \int {\dfrac{{\left[ {1 + 2f\left( x \right)} \right]'}}{{2\sqrt {1 + 2f\left( x \right)} \,}}dx} \)\( \Leftrightarrow \dfrac{2}{3}x\sqrt x  + C = \sqrt {1 + 2f\left( x \right)} \).

Theo bài ra ta có: \(f\left( 1 \right) = \dfrac{3}{2} \Leftrightarrow \dfrac{2}{3} + C = \sqrt {1 + 2.\dfrac{3}{2}} \)\( \Leftrightarrow \dfrac{2}{3} + C = 2 \Leftrightarrow C = \dfrac{4}{3}\).

Khi đó ta có:

\(\begin{array}{l}\dfrac{2}{3}x\sqrt x  + \dfrac{4}{3} = \sqrt {1 + 2f\left( x \right)} \\ \Leftrightarrow 1 + 2f\left( x \right) = {\left( {\dfrac{2}{3}x\sqrt x  + \dfrac{4}{3}} \right)^2}\\ \Leftrightarrow 1 + 2f\left( x \right) = \dfrac{4}{9}{x^3} + \dfrac{{16}}{9}x\sqrt x  + \dfrac{{16}}{9}\\ \Leftrightarrow 2f\left( x \right) = \dfrac{4}{9}{x^3} + \dfrac{{16}}{9}x\sqrt x  + \dfrac{7}{9}\\ \Leftrightarrow f\left( x \right) = \dfrac{2}{9}{x^3} + \dfrac{8}{9}x\sqrt x  + \dfrac{7}{{18}}\\ \Leftrightarrow \int\limits_1^4 {f\left( x \right)dx}  = \int\limits_1^4 {\left( {\dfrac{2}{9}{x^3} + \dfrac{8}{9}x\sqrt x  + \dfrac{7}{{18}}} \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left. {\left( {\dfrac{{{x^4}}}{{18}} + \dfrac{{16}}{{45}}{x^2}\sqrt x  + \dfrac{{7x}}{{18}}} \right)} \right|_1^4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{1222}}{{45}} - \dfrac{4}{5} = \dfrac{{1186}}{{45}}\end{array}\)

Chọn C.