Let $5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x > 0$. Then $18 \int_\limits{1}^{2} f(x) d x$ is equal to :

We have, <br/><br/>$5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x>0$ ..........(i) <br/><br/>On replacing $x$ by $\frac{1}{x}$ in (i), we get <br/><br/>$5 f\left(\frac{1}{x}\right)+4 f(x)=x+3$ ..........(ii) <br/><br/>Now, using Eq. (i) $\times 5-$ (ii) $\times 4$, we get <br/><br/>$$ \begin{aligned} & 25 f(x)-16 f(x) =\left(\frac{5}{x}+15\right)-(4 x+12) \\\\ &\Rightarrow 9 f(x) =\frac{5}{x}-4 x+3 \\\\ &\Rightarrow f(x) =\frac{1}{9}\left(\frac{5}{x}-4 x+3\right) ...........(iii) \end{aligned} $$ <br/><br/>$$ \begin{aligned} \therefore \quad & 18 \int_1^2 f(x) d x=18 \int_1^2 \frac{1}{9}\left(\frac{5}{x}-4 x+3\right) d x \text { [Using Eq. (iii)] }\\\\ = & 2 \int_1^2\left(\frac{5}{x}-4 x+3\right) d x \end{aligned} $$ <br/><br/>$$ \begin{aligned} & =2\left[5 \log _e x-4\left(\frac{x^2}{2}\right)+3 x\right]_1^2 \\\\ & =2\left[\left(5 \log _e 2-2(2)^2+3(2)\right)-\left(5 \log 1-2(1)^2+3(1)\right)\right] \\\\ & =2\left[5 \log _e 2-8+6+2-3\right] [\because \log 1=0]\\\\ & =10 \log _e 2-6 \end{aligned} $$