Answer
$ \frac{2}{x}-\frac{2}{x+1}-\frac{1}{(x+1)^2}$
Work Step by Step
1. Factor the denominator as $x^3+2x^2+x=x(x+1)^2$
2. Assume $\frac{x+2}{x^3+2x^2+x}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$
3. Evaluate the right side, we have $RHS=\frac{A(x^2+2x+1)+B(x^2+x)+Cx}{x(x+1)^2}=\frac{(A+B)x^2+(2A+B+C)x+A}{x^2+2x-3}$
4. Compare the above with the left hand side to get $\begin{cases} A+B=0 \\ 2A+B+C=1\\A=2 \end{cases}$
5. We have $B=-A=-2$ and $C=1-2A-B=-1$
6. The partial fraction decomposition is $\frac{x+2}{x^3+2x^2+x}=\frac{2}{x}-\frac{2}{x+1}-\frac{1}{(x+1)^2}$
