Answer
$\dfrac{-2}{x+1}+\dfrac{1}{(x+1)^2} $
Work Step by Step
We will write the partial decomposition as:
$\dfrac{-2x-1}{x^2+2x+1}=\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2} ~~~~(a)$
This implies that $-2x-1=A(x+1)+B$
Plug $x=0, -1$ to obtain: $A+-B=-1$ and $B=1$
After solving these equations, we get: $A=-2$ and $B=1$
So, the equation (a) becomes:
$\dfrac{-2x-1}{x^2+2x+1}=\dfrac{-2}{x+1}+\dfrac{1}{(x+1)^2} $
