Chọn: Đáp án D

Ta có:$f\left( x \right)=\frac{{{\left( x+2 \right)}^{2}}}{{{x}^{3}}}=\frac{{{x}^{2}}+4x+4}{{{x}^{3}}}=\frac{1}{x}+\frac{4}{{{x}^{2}}}+\frac{4}{{{x}^{3}}}$  $\left( x\ne 0 \right)$

$\Rightarrow F\left( x \right)=\int{f\left( x \right)dx=\int{\frac{dx}{x}+4\int{\frac{dx}{{{x}^{2}}}+4\int{\frac{dx}{{{x}^{3}}}=\ln \left| x \right|-\frac{4}{x}-\frac{2}{{{x}^{2}}}+C}}}}$

Mà $F\left( 1 \right)=6\Rightarrow C=12\Rightarrow F\left( x \right)=\ln \left| x \right|-\frac{4}{x}-\frac{2}{{{x}^{2}}}+12$