Bài giải:

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

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

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

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

Vậy đáp án đúng là: C