The function $f(x) = \int \dfrac{x+3}{x^2-9x+20}\,dx$, then $f(x)$ is

Given $f(x) = \int \dfrac{x+3}{x^2-9x+20}\,dx$

Differentiating with respect to $x$, we get:

$f'(x) = \dfrac{x+3}{x^2-9x+20} = \dfrac{x+3}{(x-4)(x-5)}$

For $f(x)$ to be decreasing, $f'(x) \le 0$.

$\dfrac{x+3}{(x-4)(x-5)} \le 0$

The critical points are $x = -3$, $x = 4$, and $x = 5$.

Using the wavy curve method to determine the sign of $f'(x)$ in different intervals:

For $x \in (5, \infty)$, $f'(x) > 0$

For $x \in (4, 5)$, $f'(x) < 0$

For $x \in (-3, 4)$, $f'(x) > 0$

For $x \in (-\infty, -3)$, $f'(x) < 0$

Therefore, $f'(x) \le 0$ for $x \in (-\infty, -3] \cup (4, 5)$.

Hence, $f(x)$ decreases on $(-\infty, -3] \cup (4, 5)$.

Answer: decreases on $(-\infty, -3] \cup (4, 5)$