Cho hàm số f(x) liên tục trên \[\mathbb{R}\backslash \left\{ 0 \right\}\] thỏa mãn \[f\left( x \right) + 4f\left( {\frac{1}{x}} \right) = 8{x^2}.\] Tính tích phân \[I = \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( x \right)}}{x}dx} .\]

Đáp án C

Đặt \[x = \frac{1}{t}\].

\[\begin{array}{l} \Rightarrow I = \int\limits_2^{\frac{1}{2}} {t.f\left( {\frac{1}{t}} \right)d\left( {\frac{1}{t}} \right)} = \int\limits_2^{\frac{1}{2}} {t.f\left( {\frac{1}{t}} \right).\frac{{ - 1}}{{{t^2}}}dt} = \int\limits_{\frac{1}{2}}^2 {\frac{1}{t}.f\left( {\frac{1}{t}} \right)dt} = \int\limits_{\frac{1}{2}}^2 {\frac{1}{x}.f\left( x \right)dx} \\ \Rightarrow 5I = \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( x \right)}}{x}dx} + 4\int\limits_{\frac{1}{2}}^2 {\frac{1}{x}.f\left( {\frac{1}{x}} \right)dx} = \int\limits_{\frac{1}{2}}^2 {\frac{1}{x}\left[ {f\left( x \right) + 4f\left( {\frac{1}{x}} \right)} \right]dx} = \int\limits_{\frac{1}{2}}^2 {\frac{1}{x}.8{x^2}dx} = 4{x^2}\left| \begin{array}{l}^2\\_{\frac{1}{2}}\end{array} \right. = 15 \Rightarrow I = 3.\end{array}\]