Lời giải:Đặt t = 2x ⇒ \({\rm{d}}x = \frac{1}{2}\,{\rm{d}}t\). Đổi cận \(\left| \begin{array}{l}
x = \frac{1}{2}\,\, \Rightarrow t = 1\\
x = \frac{3}{2}\,\, \Rightarrow t = 3
\end{array} \right.\).
Khi đó \(I=\frac{1}{2}\int\limits_{1}^{3}{f\left( \frac{2}{t} \right)\text{d}x}\).
Mà \(2f\left( 3x \right)+3f\left( \frac{2}{x} \right)=-\frac{15x}{2}\) \(\Leftrightarrow \) \(f\left( \frac{2}{x} \right)=-\frac{5x}{2}-\frac{2}{3}f\left( 3x \right)\)
Nên \(I=\frac{1}{2}\int\limits_{1}^{3}{\left[ -\frac{5x}{2}-\frac{2}{3}f\left( 3x \right) \right]\text{d}x}=-\frac{5}{4}\int\limits_{1}^{3}{x\,\text{d}x}-\frac{1}{3}\int\limits_{1}^{3}{f\left( 3x \right)\,\text{d}x}=-5-\frac{1}{3}\int\limits_{1}^{3}{f\left( 3x \right)\,\text{d}x}\) (*)
Đặt \(u=3x\) \(\Rightarrow \) \(\text{d}x=\frac{1}{3}\,\text{d}x\). Đổi cận \(\left| \begin{array}{l}
x = 1\,\, \Rightarrow u = 3\\
x = 3\, \Rightarrow t = 9
\end{array} \right.\).
Khi đó \(I=-5-\frac{1}{9}\int\limits_{3}^{9}{f\left( t \right)\,\text{d}t}=-5-\frac{k}{9}=-\frac{45+k}{9}\).
