Lời giải:\(xf\left( {{x}^{2}} \right)-f\left( 2x \right)=2{{x}^{3}}+2x\Rightarrow \int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right)-f\left( 2x \right) \right]}\text{d}x=\int\limits_{1}^{2}{\left( 2{{x}^{3}}+2x \right)\text{d}x}\)
\(\Leftrightarrow \int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x-\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x=\left( \frac{{{x}^{4}}}{2}+{{x}^{2}} \right)\left| \begin{align}
& 2 \\
& 1 \\
\end{align} \right.\Leftrightarrow \int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x-\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x=\frac{21}{2}\).
+ Tính \(\int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x\)
Đặt \(u={{x}^{2}}\Rightarrow \text{d}u=2x\text{d}x\Leftrightarrow x\text{d}x=\frac{\text{d}u}{2}\)
\(x=1\Rightarrow u=1;\,\,x=2\Rightarrow u=4\)
Suy ra \(\int\limits_{1}^{2}{\left[ xf\left( {{x}^{2}} \right) \right]}\text{d}x=\int\limits_{1}^{4}{\frac{f\left( u \right)}{2}\text{d}u}=\frac{1}{2}\int\limits_{1}^{4}{f\left( x \right)\text{d}x}\)
+ Tính \(\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x\)
Đặt \(t=2x\Rightarrow \text{d}t=2\text{d}x\Leftrightarrow \text{d}x=\frac{\text{d}t}{2}\).
\(x=1\Rightarrow t=2;\,\,x=2\Rightarrow t=4\).
Suy ra \(\int\limits_{1}^{2}{\left[ f\left( 2x \right) \right]}\text{d}x=\int\limits_{2}^{4}{\frac{f\left( t \right)}{2}\text{d}t}=\frac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x}\)
Thay vào ta được \(\frac{1}{2}\int\limits_{1}^{4}{f\left( x \right)\text{d}x}-\frac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x}=\frac{21}{2}\Leftrightarrow \frac{1}{2}\int\limits_{1}^{2}{f\left( x \right)\text{d}x}+\frac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x}-\frac{1}{2}\int\limits_{2}^{4}{f\left( x \right)\text{d}x=\frac{21}{2}}\)
\(\Leftrightarrow \frac{1}{2}\int\limits_{1}^{2}{f\left( x \right)\text{d}x}=\frac{21}{2}\Leftrightarrow \int\limits_{1}^{2}{f\left( x \right)\text{d}x}=21\).
