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 :

Question

$10 \log _{\mathrm{e}} 2+6$

$5 \log _{e} 2-3$

$10 \log _{\mathrm{e}} 2-6$

$5 \log _{\mathrm{e}} 2+3$

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