$\begin{align}  & \frac{{{x}^{2}}}{{{(x+1)}^{2}}({{x}^{2}}+1)}=\frac{A}{x+1}+\frac{B}{{{(x+1)}^{2}}}+\frac{Cx+D}{({{x}^{2}}+1)} \\ & {{x}^{2}}=A(x+1)({{x}^{2}}+1)+B({{x}^{2}}+1)+Cx+D{{(x+1)}^{2}} \\ & \text{Set }x=-1 \\ & {{(-1)}^{2}}=2B \\ & B=\frac{1}{2} \\ & \text{Expanding (i)} \\ & {{x}^{2}}=A({{x}^{3}}+{{x}^{2}}+x+1)+B({{x}^{2}}+1)+Cx+D({{x}^{2}}+2x+1) \\ & {{x}^{2}}=\left[ A{{x}^{3}}+A{{x}^{2}}+Ax+A+B{{x}^{2}}+B+C{{x}^{3}}+2C{{x}^{2}}+Cx+D{{x}^{2}}+2Dx+D \right] \\ & \text{Comparing terms} \\ & {{x}^{3}}-term \\ & A+C=0---(i) \\ & {{x}^{2}}-term \\ & A+B+2C+D=1---(ii) \\ & x-term \\ & A+C+2D=0---(iii) \\ & \text{constant term} \\ & A+B+D=0---(iv) \\ & \text{Substitute }\tfrac{1}{2}\text{ for }B\text{ in eqn }(iv)\text{ } \\ & A+\tfrac{1}{2}+D=0 \\ & A+D=-\tfrac{1}{2}---(v) \\ & \text{Substitute }\tfrac{1}{2}\text{ for }B\text{  and }-\tfrac{1}{2}\text{ for }(A+D)\text{ in eqn }(ii) \\ & -\tfrac{1}{2}+\tfrac{1}{2}+2C=1 \\ & C=\tfrac{1}{2} \\ & \text{Substitute }\tfrac{1}{2}\text{ for }C\text{ in }(i) \\ & A+\tfrac{1}{2}=0 \\ & A=-\tfrac{1}{2} \\ & \text{Substitute }-\tfrac{1}{2}\text{ for }A\text{ in equation }(v) \\ & -\tfrac{1}{2}+D=-\tfrac{1}{2},\text{   }D=0 \\ & \text{The partial fraction will be} \\ & \frac{{{x}^{2}}}{{{(x+1)}^{2}}({{x}^{2}}+1)}=\frac{-\tfrac{1}{2}}{x+1}+\frac{\tfrac{1}{2}}{{{(x+1)}^{2}}}+\frac{\tfrac{1}{2}x}{({{x}^{2}}+1)} \\ & \frac{{{x}^{2}}}{{{(x+1)}^{2}}({{x}^{2}}+1)}=\frac{1}{2}\left[ \frac{-1}{x+1}+\frac{1}{{{(x+1)}^{2}}}+\frac{1}{{{x}^{2}}+1} \right] \\\end{align}$