Given:

$5f\left(x\right)+4f\left(\frac{1}{x}\right)=\frac{1}{x}+3 ....\left(1\right)$

Replace $x\to \frac{1}{x}$

$5f\left(\frac{1}{x}\right)+4f\left(x\right)=x+3 ....\left(2\right)$

Eliminating $f\left(\frac{1}{x}\right)$ from $\left(1\right) \& \left(2\right)$, we get

$9f\left(x\right)=\frac{5}{x}+15-4x-12$

$\Rightarrow 9f\left(x\right)=\frac{5}{x}-4x+3$

$\Rightarrow 18f\left(x\right)=\frac{10}{x}-8x+6$

$\Rightarrow 18{\int }_1^{2}f\left(x\right)dx={\int }_1^2\left(\frac{10}{x}-8x+6\right)dx$

$\Rightarrow 18{\int }_1^{2}f\left(x\right)dx={\left[10\log x-4x^2+6x\right]}_1^2$

$\Rightarrow 18{\int }_1^{2}f\left(x\right)dx=\left[\left(10\log 2-4\right)-\left(10\log 1+2\right)\right]$

$\Rightarrow 18{\int }_1^{2}f\left(x\right)dx=10{\log }_{e}2-6$

Hence this is the correct option.