Let a _ n = ∫ _ - 1 ^ n (1 + (x)/(2) + x ^ 2 3 + + x ^ n - 1 n )dx...

Given ${a}_{n} = {∫}_{- 1}^{n}\left(1 + \dfrac{x}{2} + \dfrac{{x}^{2}}{3} + \ldots . + \dfrac{{x}^{n - 1}}{n}\right)dx$

$= {\left[x + \dfrac{{x}^{2}}{{2}^{2}} + \dfrac{{x}^{3}}{{3}^{2}} + \ldots \ldots + \dfrac{{x}^{n}}{{n}^{2}}\right]}_{- 1}^{n}$

i.e. ${a}_{n} = \dfrac{n + 1}{{1}^{2}} + \dfrac{{n}^{2} - 1}{{2}^{2}} + \dfrac{{n}^{3} + 1}{{3}^{2}} + \dfrac{{n}^{4} - 1}{{4}^{2}} + \ldots + \dfrac{{n}^{n} + {\left(- 1\right)}^{n + 1}}{{n}^{2}}$

Here ${a}_{1} = 2, {a}_{2} = \dfrac{2 + 1}{1} + \dfrac{{2}^{2} - 1}{2} = \dfrac{9}{2},$

${a}_{3} = 4 + 2 + \dfrac{28}{9} = \dfrac{100}{9},$

${a}_{4} = 5 + \dfrac{15}{4} + \dfrac{65}{9} + \dfrac{255}{16} > 31 .</p><p>$∵ {a}_{n} \in \left(2, 30\right)$, so the required set is \left\{2, 3\right\}$.

$∴$ sum of all elements $= 5$.