Cho hàm số y = f(x) liên tục trên đoạn $\left[ \dfrac{1}{2};2...

Ta có:
$f\left( x \right)+2f\left( \dfrac{1}{x} \right)=3x$, chia cả 2 vế cho x ta được $\dfrac{f\left( x \right)}{x}+2\dfrac{f\left( \dfrac{1}{x} \right)}{x}=3$
Lấy tích phân 2 vế
$\int\limits_{\dfrac{1}{2}}^{2}{\left[ \dfrac{f\left( x \right)}{x}+2\dfrac{f\left( \dfrac{1}{x} \right)}{x} \right]dx}=\int\limits_{\dfrac{1}{2}}^{2}{3dx}$
$\Leftrightarrow \int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{x}dx+2\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( \dfrac{1}{x} \right)}{x}dx}=3x\left| \begin{aligned}
& 2 \\
& \dfrac{1}{2} \\
\end{aligned} \right.=\dfrac{9}{2}}$
Xét $\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( \dfrac{1}{x} \right)}{x}dx}$ : Đặt $\dfrac{1}{x}=t\Rightarrow -\dfrac{1}{{{x}^{2}}}dx=dt\Rightarrow dx=-\dfrac{dt}{{{t}^{2}}}.$ Đổi cận $\left\{ \begin{aligned}
& x=\dfrac{1}{2}\Rightarrow t=2 \\
& x=2\Rightarrow t=\dfrac{1}{2} \\
\end{aligned} \right..$
Khi đó $\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( \dfrac{1}{x} \right)}{x}dx}=-\int\limits_{2}^{\dfrac{1}{2}}{\dfrac{t.f\left( t \right)}{{{t}^{2}}}dt}\Rightarrow \int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( \dfrac{1}{x} \right)}{x}dx}=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( t \right)}{t}dt=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{x}dx.}}$
Thay vào tích phân ban đầu ta được
$3\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{x}dx}=\dfrac{9}{2}\Rightarrow \int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{x}dx}=\dfrac{3}{2}.$