Solution:
Let the two integers be $a$ and $b$ with $a > b$.

Let $S = a + b$ and $D = a - b$.

We are told:

$$\frac{S}{D} = 3$$

and

$$S \cdot D = 300.$$

From the first equation, $S = 3D$.

Substitute into the second equation:

$$(3D) \cdot D = 300 \\ 3D^2 = 300 \\ D^2 = 100 \\ D = 10 \text{ or } D = -10.$$

If $D = 10$, then $S = 3 \times 10 = 30$.
If $D = -10$, then $S = 3 \times (-10) = -30$.

Recall $a = \frac{S + D}{2}$, $b = \frac{S - D}{2}$.

Case 1: $D = 10$, $S = 30$

$$a = \frac{30 + 10}{2} = 20 \\ b = \frac{30 - 10}{2} = 10$$

Case 2: $D = -10$, $S = -30$

$$a = \frac{-30 + (-10)}{2} = \frac{-40}{2} = -20 \\ b = \frac{-30 - (-10)}{2} = \frac{-20}{2} = -10$$

Thus, the integers are $20$ and $10$, or $-20$ and $-10$.