Let \( f(x)+2f\!\left(\dfrac{1}{x}\right)=x^2+5 \) and \( 2g(x)-3g\!\left(\dfrac{1}{2}\right)=x,\; x>0 \). If \( \alpha=\displaystyle \int_{1}^{2} f(x)\,dx \) and \( \beta=\displaystyle \int_{1}^{2} g(x)\,dx \), then the value of \( 9\alpha+\beta \) is:
