Let $$a_n = \displaystyle\int_{-1}^{n} \left(1 + \dfrac{x}{2} + \dfrac{x^2}{3} + \ldots + \dfrac{x^{n-1}}{n}\right) dx$$ for every $$n \in \mathbb{N}$$. Then the sum of all the elements of the set $$\{n \in \mathbb{N} : a_n \in (2, 30)\}$$ is ______.

Consider $$a_n = \displaystyle\int_{-1}^{n}\left(1 + \frac{x}{2} + \frac{x^2}{3} + \ldots + \frac{x^{n-1}}{n}\right)dx$$ for every $$n \in \mathbb{N}$$.

Integrating term by term and using that the integrand is $$\displaystyle\sum_{k=1}^{n} \frac{x^{k-1}}{k}\,$$, we obtain

$$a_n = \sum_{k=1}^{n} \frac{1}{k}\int_{-1}^{n} x^{k-1}\,dx = \sum_{k=1}^{n} \frac{1}{k}\cdot\frac{x^k}{k}\Bigg|_{-1}^{n} = \sum_{k=1}^{n} \frac{n^k - (-1)^k}{k^2}$$

For $$n = 1$$,

$$a_1 = \frac{1^1 - (-1)^1}{1^2} = \frac{1 + 1}{1} = 2$$

Since $$2 \notin (2, 30)$$ (the interval is open), $$n = 1$$ is excluded.

For $$n = 2$$,

$$a_2 = \frac{2^1 - (-1)}{1} + \frac{2^2 - 1}{4} = \frac{3}{1} + \frac{3}{4} = 3 + 0.75 = 3.75$$

Since $$3.75 \in (2, 30)$$, $$n = 2$$ is included.

For $$n = 3$$,

$$a_3 = \frac{3^1 + 1}{1} + \frac{3^2 - 1}{4} + \frac{3^3 + 1}{9} = \frac{4}{1} + \frac{8}{4} + \frac{28}{9} = 4 + 2 + \frac{28}{9} = 6 + \frac{28}{9} = \frac{82}{9} \approx 9.11$$

Since $$9.11 \in (2, 30)$$, $$n = 3$$ is included.

For $$n = 4$$,

$$a_4 = \frac{4+1}{1} + \frac{16-1}{4} + \frac{64+1}{9} + \frac{256-1}{16}$$ $$= 5 + \frac{15}{4} + \frac{65}{9} + \frac{255}{16} = 5 + 3.75 + 7.22 + 15.94 = 31.91$$

Since $$31.91 > 30$$, $$n = 4$$ is excluded.

For $$n \geq 4$$, the dominant term $$\frac{n^n}{n^2}$$ grows extremely fast, so $$a_n > 30$$ for all $$n \geq 4$$.

The set of natural numbers for which $$a_n \in (2, 30)$$ is $$\{2, 3\}$$ and its sum is

$$\text{Sum} = 2 + 3 = 5$$

The answer is $$\boxed{5}$$.