The complete set of values of x in which $f(x)=2\log_e(x-2)-x^2+4x+1$ increases, is:

(2, 3)

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

Domain is : x > 2 (as log takes positive inputs)

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

For f(x) to increase f'(x) > 0 $⇒\frac{-2(x-3)(x-1)}{(x-2)}>0⇒\frac{(x-3)(x-1)}{(x-2)}<0$

But x > 2  ⇒ x - 2 > 0 and x - 1 > 0  ⇒ (x - 3) < 0 ⇒ x < 3  

Hence x ∈ (2, 3)