It is given that \(y=(x^2+1)(2x-3)^{1/2}\).

(i) Show that

\(\frac{dy}{dx}=\frac{Px^2+Qx+1}{(2x-3)^{1/2}},\)

where \(P\) and \(Q\) are integers.

(ii) Hence find the equation of the normal to the curve at the point where \(x=2\), giving your answer in the form \(ax+by+c=0\), where \(a\), \(b\) and \(c\) are integers.