The function $f(x)=2\log (x-2)-x^2+4x+1$ increases on the interval:

(-∞ ,1) ∪ (2, 3) 

$f(x)=2\log (x-2)-x^2+4x+1$

$⇒f'(x)=\frac{2}{x-2}-2x+4⇒f'(x)=2[\frac{1-(x-2)^2}{x-2}]=-2\frac{(x-1)(x-3)}{x-2}$

$⇒f'(x)=\frac{2(x-1)(x-3)(x-2)}{(x-2)^2}$

$∴f'(x)>0⇒-2(x-1)(x-3)(x-2)>0$

$⇒(x-1)(x-3)(x-2)<0$

$⇒x∈(-∞ ,1) ∪ (2, 3) $

Thus, f (x) is increasing on (-∞ ,1) ∪ (2, 3)