Let f(x)+2 f(1/x)=x^2+5 and 2 g(x)-3 g (1/2)=x, x>0. If α=∫_1^2 f(x) d x, and β=∫_1^2 g (x) d x, then the value of 9 α+β is:

Let $\displaystyle f(x)+2 f\left(\frac{1}{x}\right)=x^2+5$ and $\displaystyle 2 g(x)-3 \mathrm{~g}\left(\frac{1}{2}\right)=x, x>0$. If $\displaystyle \alpha=\int_1^2 f(x) \mathrm{d} x$, and $\displaystyle \beta=\int_1^2 \mathrm{~g}(x) \mathrm{d} x$, then the value of $\displaystyle 9 \alpha+\beta$ is :