Bài giải:

$\frac{{{x}^{2}}+2}{{{x}^{3}}-1}+\frac{2}{{{x}^{2}}+x+1}+\frac{1}{1-x}$= $\frac{{{x}^{2}}+2}{(x-1)({{x}^{2}}+x+1)}+\frac{2}{{{x}^{2}}+x+1}+\frac{-1}{x-1}$

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

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

Vậy đáp án đúng là: D