- Chuyển \(2{\left[ {f\left( x \right)} \right]^2}\) sang VT, chia cả 2 vế cho \(2f\left( x \right)\).- Chia cả 2 vế cho \({x^2}\).- Lấy tích phân từ 1 đến 4 hai vế.Giải chi tiết:Theo bài ra ta có:\(\begin{array}{l}\,\,\,\,\,\,2x.f\left( x \right).f'\left( x \right) = {x^3} + 2{\left[ {f\left( x \right)} \right]^2}\,\,\forall x \in \left[ {1;4} \right]\\ \Leftrightarrow 2x.f\left( x \right).f'\left( x \right) - 2{\left[ {f\left( x \right)} \right]^2} = {x^3}\,\,\forall x \in \left[ {1;4} \right]\\ \Leftrightarrow xf'\left( x \right) - f\left( x \right) = \dfrac{{{x^3}}}{{2f\left( x \right)}}\,\,\forall x \in \left[ {1;4} \right]\\ \Leftrightarrow \dfrac{{xf'\left( x \right) - f\left( x \right)}}{{{x^2}}} = \dfrac{x}{{2f\left( x \right)}}\,\,\forall x \in \left[ {1;4} \right]\\ \Leftrightarrow \dfrac{{f'\left( x \right).x - f\left( x \right).x'}}{{{x^2}}} = \dfrac{x}{{2f\left( x \right)}}\,\,\forall x \in \left[ {1;4} \right]\\ \Leftrightarrow \left( {\dfrac{{f\left( x \right)}}{x}} \right)' = \dfrac{x}{{2f\left( x \right)}}\,\,\forall x \in \left[ {1;4} \right]\end{array}\)Lấy tích phân từ 1 đến 4 hai vế ta được: \(\int\limits_1^4 {\left( {\dfrac{{f\left( x \right)}}{x}} \right)'dx} = \dfrac{1}{2}\int\limits_1^4 {\dfrac{x}{{f\left( x \right)}}dx} \).\( \Leftrightarrow \left. {\dfrac{{f\left( x \right)}}{x}} \right|_1^4 = \dfrac{1}{2}\int\limits_1^4 {\dfrac{x}{{f\left( x \right)}}dx} \Leftrightarrow \int\limits_1^4 {\dfrac{x}{{f\left( x \right)}}dx} = 2\left( {\dfrac{1}{4}f\left( 4 \right) - f\left( 1 \right)} \right) = \dfrac{1}{2}.8 - 2.1 = 2\).Chọn A
