Let \(\displaystyle 5f(x)+4f\!\left(\frac{1}{x}\right)=\frac{1}{x}+3,\, x>0 \). Then \(\displaystyle 18\!\int_{1}^{2} f(x)\,dx \) is equal to:
\( 5\log_{e}2+3 \)
\( 10\log_{e}2+6 \)
\( 10\log_{e}2-6 \)
\( 5\log_{e}2-3 \)