Bài giải:

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

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

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

 Vậy đáp án đúng là: C