Answer
\[\frac{-3}{x+2}+\frac{3}{x+1}+\frac{-1}{(x+1)^2}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{2x+1}{(x+2)(x+1)^2}=\frac{A}{x+2}+\frac{B}{x+1}+\frac{C}{(x+1)^2}\;\;\;...(1)\]
\[2x+1=A(x+1)^2+B(x+1)(x+2)+C(x+2)\]
\[2x+1=A(x^2+2x+1)+B(x^2+3x+2)+C(x+2)\]
\[2x+1=(A+B)x^2+(2A+3B+C)x+(A+2B+2C)\]
Comparing like coefficients
\[ A+B=0\Rightarrow B=-A\;\;\;...(2)\]
\[2A+3B+C=2\]
From (2)
\[-A+C=2\;\;\;...(3)\]
Also $$A+2B+2C=1$$
From (2)
$$-A+2C=1\;\;\;...(4)$$
Subtract (4) from (3)
\[-C=1\Rightarrow C=-1\]
From (3) \[A=-3\]
From (2) \[B=3\]
From (1)
\[\frac{2x+1}{(x+2)(x+1)^2}=\frac{-3}{x+2}+\frac{3}{x+1}+\frac{-1}{(x+1)^2}\]
Hence, \[\frac{2x+1}{(x+2)(x+1)^2}=\frac{-3}{x+2}+\frac{3}{x+1}+\frac{-1}{(x+1)^2}.\]
