Chọn: Đáp án B

Ta có: $f\left( x \right)=\frac{1}{{{x}^{3}}+{{x}^{5}}}=\frac{\left( 1+{{x}^{2}} \right)-{{x}^{2}}}{{{x}^{3}}\left( 1+{{x}^{2}} \right)}=\frac{1}{{{x}^{3}}}-\frac{1}{x\left( 1+{{x}^{2}} \right)}=\frac{1}{{{x}^{3}}}-\frac{\left( 1+{{x}^{2}} \right)-{{x}^{2}}}{x\left( 1+{{x}^{2}} \right)}=\frac{1}{{{x}^{3}}}-\frac{1}{x}+\frac{x}{1+{{x}^{2}}}$

Vậy $\int{f\left( x \right)}dx=\int{\frac{dx}{{{x}^{3}}}}-\int{\frac{dx}{x}}+\int{\frac{xdx}{1+{{x}^{2}}}}=-\frac{1}{2{{x}^{2}}}-\ln \left| x \right|+\frac{1}{2}\ln \left( {{x}^{2}}+1 \right)+C$