$2 f(x)-3 f\left(\frac{1}{x}\right)=x$
...(1) and replacing $x$ by $\frac{1}{x}$, we get
$$
2 f\left(\frac{1}{x}\right)-3 f(x)=\frac{1}{x}
$$
[ $2 \times$ equation (1) $]+[3 \times$ equation (2) $]$ gives,
$$
\begin{aligned}
& 4 f(x)-9 f(x)=2 x+\frac{3}{x} \quad \Rightarrow-5 f(x)=2 x+\frac{3}{x} \\
& \therefore f(x)=\frac{-2}{5} x-\frac{3}{5 x}
\end{aligned}
$$


$$
\begin{aligned}
& \therefore \int_1^{\mathrm{e}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_1^{\mathrm{e}}\left(\frac{-2}{5} \mathrm{x}-\frac{3}{5 \mathrm{x}}\right) \mathrm{d} x=\frac{-2}{5} \int_1^{\mathrm{e}} \mathrm{x} d \mathrm{x}-\frac{3}{5} \int_1^{\mathrm{e}} \frac{1}{\mathrm{x}} \mathrm{dx} \\
& =\frac{-2}{5}\left[\frac{\mathrm{x}^2}{2}\right]_1^{\mathrm{e}}-\frac{3}{5}[\log \mathrm{x}]_1^{\mathrm{e}}=\frac{-1}{5}\left(\mathrm{e}^2-1\right)-\frac{3}{5}(\log \mathrm{e}-\log 1)=\frac{-1}{5} \mathrm{e}^2+\frac{1}{5}-\frac{3}{5} \\
& =-\left(\frac{2+\mathrm{e}^2}{5}\right)
\end{aligned}
$$