The function $f(x) = 2\log_e(x-2) - x^2 + 4x + 1, (x > 2)$ is increasing on the interval:

A function is increasing when its derivative is positive.

Given: $f(x) = 2\log_e(x-2) - x^2 + 4x + 1$

Differentiating term by term:

$\frac{d}{dx}[2\log_e(x-2)] = \frac{2}{x-2}$

$\frac{d}{dx}[-x^2] = -2x$

$\frac{d}{dx}[4x] = 4$

$\frac{d}{dx}[1] = 0$

Therefore:

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

---

For the function to be increasing:

$f'(x) > 0$

$\frac{2}{x-2} - 2x + 4 > 0$

$\frac{2}{x-2} > 2x - 4$

$\frac{2}{x-2} > 2(x - 2)$

$\frac{1}{x-2} > (x - 2)$

---

Since $x > 2$, we have $(x-2) > 0$.

Multiplying both sides by $(x-2)$:

$1 > (x-2)^2$

$(x-2)^2 < 1$

---

Taking the square root:

$|x-2| < 1$

$-1 < x-2 < 1$

$1 < x < 3$

---

The original function has domain $x > 2$.

Combining $x > 2$ and $1 < x < 3$:

$2 < x < 3$

Therefore, the function is increasing on the interval $(2, 3)$.