Evaluation of $\sum_{n=0}^{\infty}\frac{1}{(n^4+n^2+1)n!}$

We can rewrite the prefactor as $$\frac{1}{n^4+n^2+1}=\frac{1}{(n^2-n+1)(n^2+n+1)}=a_{n+1}-na_n+\frac12,$$ with $\displaystyle a_n=\frac{n}{2(n^2-n+1)}$. Now it is easy to understand that $a_n$'s give a sum that telescopes to $0$, so that we are left with $$\frac12\sum_{n=0}^{\infty}\frac{1}{n!}=\frac e2.$$

Added on request of OP: $$\sum_{n=0}^{\infty}\frac{-na_n+a_{n+1}}{n!}=-\frac{0\cdot a_0}{0!}+{\color{red}{\frac{a_{1}}{0!}-\frac{1\cdot a_1}{1!}}}+{\color{blue}{\frac{a_2}{1!}-\frac{2\cdot a_2}{2!}}}+{\color{magenta}{\frac{a_3}{2!}-\frac{3\cdot a_3}{3!}}}+\frac{a_4}{3!}+\ldots=0.$$