1. We are given $5f(x)+4f\left(\frac{1}{x}\right)=\frac{1}{x}+3, \qquad x>0.$ We need to find $18\int_1^2 f(x)\,dx.$

2. Write the same relation with $x$ replaced by $\frac{1}{x}$: $5f\left(\frac{1}{x}\right)+4f(x)=x+3.$

So we have the system: $5f(x)+4f\left(\frac{1}{x}\right)=\frac{1}{x}+3 \quad \text{...(1)}$ $4f(x)+5f\left(\frac{1}{x}\right)=x+3 \quad \text{...(2)}$

3. Let $a=f(x), \qquad b=f\left(\frac{1}{x}\right).$ Then $5a+4b=\frac{1}{x}+3,$ $4a+5b=x+3.$

Multiply the first by $5$ and the second by $-4$: $25a+20b=\frac{5}{x}+15,$ $-16a-20b=-4x-12.$ Adding, $9a=\frac{5}{x}+3-4x.$ Hence $f(x)=a=\frac{1}{9}\left(\frac{5}{x}+3-4x\right).$

4. Now integrate from $1$ to $2$: $\int_1^2 f(x)\,dx=\frac{1}{9}\int_1^2 \left(\frac{5}{x}+3-4x\right)dx.$

Compute each part: $\int_1^2 \frac{5}{x}\,dx=5\ln 2,$ $\int_1^2 3\,dx=3,$ $\int_1^2 4x\,dx=2x^2\Big|_1^2=2(4)-2(1)=6.$

So $\int_1^2 f(x)\,dx=\frac{1}{9}(5\ln 2+3-6)=\frac{1}{9}(5\ln 2-3).$

5. Therefore, $18\int_1^2 f(x)\,dx=18\cdot \frac{1}{9}(5\ln 2-3)=2(5\ln 2-3).$ Thus $18\int_1^2 f(x)\,dx=10\ln 2-6.$

6. Compare with options:

• A: $10\ln 2+6$
• B: $5\ln 2-3$
• C: $10\ln 2-6$
• D: $5\ln 2+3$

Hence the correct option is $\boxed{\text{C } (10\ln 2-6)}.$