If \(2 f(x)-3 f\left(\frac{1}{x}\right)=x\), then \(\int_1^e f(x) d x=\)

\(2 f(x)-3 f\left(\frac{1}{x}\right)=x\)
...(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,
\(
\begin{aligned}
& 4 f(x)-9 f(x)=2 x+\frac{3}{x} \quad \Rightarrow-5 f(x)=2 x+\frac{3}{x} \\
& \therefore f(x)=\frac{-2}{5} x-\frac{3}{5 x}
\end{aligned}
\)
\(\therefore \int_1^{\mathrm{e}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_1^{\mathrm{e}}\left(\frac{-2}{5} \mathrm{x}-\frac{3}{5 \mathrm{x}}\right) \mathrm{d} x=\frac{-2}{5} \int_1^{\mathrm{e}} \mathrm{x} d \mathrm{x}-\) \(\frac{3}{5} \int_1^{\mathrm{e}} \frac{1}{\mathrm{x}} \mathrm{dx} \)
\( =\frac{-2}{5}\left[\frac{\mathrm{x}^2}{2}\right]_1^{\mathrm{e}}-\frac{3}{5}[\log \mathrm{x}]_1^{\mathrm{e}}=\frac{-1}{5}\left(\mathrm{e}^2-1\right)-\frac{3}{5}(\log \mathrm{e}\) \(-\log 1)=\frac{-1}{5} \mathrm{e}^2+\frac{1}{5}-\frac{3}{5} \)
\( =-\left(\frac{2+\mathrm{e}^2}{5}\right)\)