$\begin{align}  & \frac{7{{x}^{2}}+12x+4}{(x+2){{(x+1)}^{2}}}=\frac{A}{x+2}+\frac{B}{x+1}+\frac{C}{{{(x+1)}^{2}}} \\  & 7{{x}^{2}}+12x+4=A{{(x+1)}^{2}}+B(x+2)(x+1)+C(x+2) \\  & \text{Let }x=-2 \\  & A=28-24+4 \\  & A=8 \\  & \text{Let }x=-1 \\  & C=7-12+4=-1 \\  & \text{Let }x=0 \\  & 4=A+2B+2C \\  & \text{Substitute }A=8,\text{ }C=-1\text{ into }A+2B+2C=4 \\  & 8+2B-2=4 \\  & 2B=-2 \\  & B=-1 \\  & \frac{7{{x}^{2}}+12x+4}{(x+2){{(x+1)}^{2}}}=\frac{8}{x+2}-\frac{1}{x+1}-\frac{1}{{{(x+1)}^{2}}} \\  & \frac{8}{x+2}-\frac{1}{x+1}-\frac{1}{{{(x+1)}^{2}}}=8{{(x+2)}^{-1}}-{{(x+1)}^{-1}}-{{(x+1)}^{-2}} \\  & \text{The expansion gives neglecting }{{x}^{3}}\text{ and higher terms} \\  & =8\left[ {{2}^{-1}}{{(1+\tfrac{x}{2})}^{-1}} \right]-{{(x+1)}^{-1}}-{{(x+1)}^{-2}} \\  & =4{{(1+\tfrac{x}{2})}^{-1}}-{{(x+1)}^{-1}}-{{(x+1)}^{-2}} \\  & =4\left[ 1-\frac{x}{2}+\frac{{{x}^{2}}}{4}+--- \right]-\left[ 1-x+{{x}^{2}} \right]-\left[ 1-2x+3{{x}^{2}} \right] \\  & =4-2x+{{x}^{2}}-1+x-{{x}^{2}}+2x-3{{x}^{2}}+---- \\  & =2+x-3{{x}^{2}}+----- \\  & \frac{7{{x}^{2}}+12x+4}{(x+2){{(x+1)}^{2}}}=\frac{8}{x+2}-\frac{1}{x+1}-\frac{1}{{{(x+1)}^{2}}} \\  & \frac{8}{x+2}-\frac{1}{x+1}-\frac{1}{{{(x+1)}^{2}}}=8{{(x+2)}^{-1}}-{{(x+1)}^{-1}}-{{(x+1)}^{-2}} \\  & \text{The expansion gives neglecting }{{x}^{3}}\text{ and higher terms} \\  & =8\left[ {{2}^{-1}}{{(1+\tfrac{x}{2})}^{-1}} \right]-{{(x+1)}^{-1}}-{{(x+1)}^{-2}} \\  & =4{{(1+\tfrac{x}{2})}^{-1}}-{{(x+1)}^{-1}}-{{(x+1)}^{-2}} \\  & =4\left[ 1-\frac{x}{2}+\frac{{{x}^{2}}}{4}+--- \right]-\left[ 1-x+{{x}^{2}} \right]-\left[ 1-2x+3{{x}^{2}} \right] \\  & =4-2x+{{x}^{2}}-1+x-{{x}^{2}}+2x-3{{x}^{2}}+---- \\  & =2+x-3{{x}^{2}}+----- \\ \end{align}$