$\begin{align}  & \frac{{{x}^{2}}-5}{{{(x+1)}^{2}}({{x}^{2}}+1)}=\frac{A}{x+1}+\frac{B}{{{(x+1)}^{2}}}+\frac{Cx+D}{({{x}^{2}}+1)} \\ & {{x}^{2}}-5=A(x+1)({{x}^{2}}+1)+B({{x}^{2}}+1)+(Cx+D){{(x+1)}^{2}} \\ & \text{Let }x=-1 \\ & 1-5=2B \\ & B=-2 \\ & \text{Expanding equation }(i) \\ & {{x}^{2}}-5=\left[ A({{x}^{3}}+{{x}^{2}}+x+1)+B({{x}^{2}}+1)+(Cx+D)({{x}^{2}}+2x+1) \right] \\ & {{x}^{2}}-5=\left[ A{{x}^{3}}+A{{x}^{2}}+Ax+A+B{{x}^{2}}+B+C{{x}^{3}}+2C{{x}^{2}}+Cx+D{{x}^{2}}+2Dx+D \right] \\ & \text{Comparing identities} \\ & {{x}^{3}}-terms \\ & A{{x}^{3}}+C{{x}^{3}}=0{{x}^{3}} \\ & A+C=0---(i) \\ & {{x}^{2}}-terms \\ & A{{x}^{2}}+B{{x}^{2}}+2C{{x}^{2}}+D{{x}^{2}}={{x}^{2}} \\ & A+B+2C+D=1---(ii) \\ & x-term \\ & Ax+Cx+2Dx=0x \\ & A+C+2D=0---(iii) \\ & \text{constant term} \\ & A+B+D=-5---(iv) \\ & \text{Substitute for }-2\text{ for }B\text{ in  }(iv) \\ & A-2+D=-5 \\ & A+D=-3-----(v) \\ & \text{substitute }-3\text{ for }A+D\text{ and }-2\text{ for }B\text{ in eqn (ii)} \\ & -3+2C-2=1 \\ & 2C=6 \\ & C=3 \\ & \text{Substitute 3 for }C\text{ in eqn(i)} \\ & A+3=0 \\ & A=-3 \\ & \text{subsitiute }A=-3\text{ in }eqn(v) \\ & -3+D=-3 \\ & D=0 \\ & \text{The partial fraction becomes } \\ & \frac{{{x}^{2}}-5}{{{(x+1)}^{2}}({{x}^{2}}+1)}=\frac{-3}{x+1}-\frac{2}{{{(x+1)}^{2}}}+\frac{3x}{{{x}^{2}}+1} \\\end{align}$   $\begin{align}  & \frac{3x+2}{(x-3)(x+2)}=\frac{A}{(x-3)}+\frac{B}{(x+2)} \\ & 3x+2=A(x+2)+B(x-3) \\ & \text{set }x=3 \\ & 3(3)+2=A(3+2) \\ & 11=5A \\ & A=\frac{11}{5}----(i) \\ & \text{set }x=-2 \\ & 3(-2)+2=B(-2-3) \\ & -6+2=-5B \\ & B=\frac{4}{5} \\ & \text{The partial fraction becomes} \\ & \frac{3x+2}{(x-3)(x+2)}=\frac{\tfrac{11}{5}}{(x-3)}+\frac{\tfrac{4}{5}}{(x+2)} \\ & \frac{3x+2}{(x-3)(x+2)}=\frac{1}{5}\left( \frac{11}{(x-3)}+\frac{4}{(x+2)} \right) \\\end{align}$