Cho hàm số \(y = f\left( x \right)\) liên tục trên...

A

- Hướng dẫn giải

Phương pháp giải:

Tích phân hai vế của \(f\left( x \right) = \dfrac{{f\left( {2\sqrt x  - 1} \right)}}{{\sqrt x }} + \dfrac{{\ln x}}{x}\).

Giải chi tiết:

\(\begin{array}{l}f\left( x \right) = \dfrac{{f\left( {2\sqrt x  - 1} \right)}}{{\sqrt x }} + \dfrac{{\ln x}}{x} \Rightarrow \int\limits_1^4 {f\left( x \right)} dx = \int\limits_1^4 {\left( {\dfrac{{f\left( {2\sqrt x  - 1} \right)}}{{\sqrt x }} + \dfrac{{\ln x}}{x}} \right)} dx\\ \Leftrightarrow \int\limits_1^4 {f\left( x \right)} dx = \int\limits_1^4 {\dfrac{{f\left( {2\sqrt x  - 1} \right)}}{{\sqrt x }}} dx + \int\limits_1^4 {\dfrac{{\ln x}}{x}} dx\end{array}\)

Tính \({I_1} = \int\limits_1^4 {\dfrac{{f\left( {2\sqrt x  - 1} \right)}}{{\sqrt x }}} dx = \int\limits_1^4 {f\left( {2\sqrt x  - 1} \right)\,} d\left( {2\sqrt x  - 1} \right) = \int\limits_1^3 {f\left( t \right)\,} dt = \int\limits_1^3 {f\left( x \right)\,} dx\)

Tính \({I_2} = \int\limits_1^4 {\dfrac{{\ln x}}{x}} dx = \int\limits_1^4 {\ln x} d\left( {\ln x} \right) = \left. {\dfrac{{{{\left( {\ln x} \right)}^2}}}{2}} \right|_1^4 = \dfrac{{{{\left( {\ln 4} \right)}^2}}}{2} = 2{\ln ^2}2\)\( \Rightarrow \int\limits_1^4 {f\left( x \right)} dx = \int\limits_1^3 {f\left( x \right)} dx + 2{\ln ^2}2\)

\( \Leftrightarrow \int\limits_1^4 {f\left( x \right)} dx - \int\limits_1^3 {f\left( x \right)} dx = 2{\ln ^2}2 \Leftrightarrow \int\limits_1^4 {f\left( x \right)} dx + \int\limits_3^1 {f\left( x \right)} dx = 2{\ln ^2}2 \Leftrightarrow \int\limits_3^4 {f\left( x \right)} dx = 2{\ln ^2}2\)

Vậy \(I=\int\limits_{3}^{4}{f\left( x \right)dx}\)\( = 2{\ln ^2}2\).

Chọn: A