Đặt $x=-t\Rightarrow dx=-dt$.

Đổi cận $x=1\Rightarrow t=-1;x=-1\Rightarrow t=1$.

Suy ra

$I=\int_{-1}^{1}{\frac{{{x}^{4}}}{1+{{2}^{x}}}dx}=-\int_{1}^{-1}{\frac{{{t}^{4}}}{1+{{2}^{-t}}}dt}=\int_{-1}^{1}{\frac{{{t}^{4}}dt}{1+{{2}^{-t}}}}$

$=\int_{-1}^{1}{\frac{{{t}^{4}}{{.2}^{t}}}{1+{{2}^{t}}}dt}=\int_{-1}^{1}{\frac{{{t}^{4}}{{.2}^{t}}+{{t}^{4}}-{{t}^{4}}}{1+{{2}^{t}}}}dt$

Do đó $I=\int_{-1}^{1}{\frac{{{x}^{4}}}{1+{{2}^{x}}}dx=\frac{1}{2}\int_{-1}^{1}{{{t}^{4}}dt}=\left. \frac{1}{2}.\frac{{{t}^{5}}}{5} \right|}_{-1}^{1}=\frac{1}{5}$.

Vậy $\int_{-1}^{1}{\frac{{{x}^{4}}}{1+{{2}^{x}}}dx=\frac{1}{5}}$. Ta chọn phương án C.