Split into partial fractions $\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}$ This is from "Calculus Made Easy", Exercises 10, Question 15 (page 147).  I've worked this one over and over and still haven't made progress.
This is my initial setup:
$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2} = \frac{A}{(x+2)} + \frac{Bx+C}{(x^2+x+1)} + \frac{Dx+E}{(x^2+x+1)^2}$$
then:
$$3x^2+2x+1 = A(x^2+x+1)^2 + (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
and I can solve for A by setting $x=-2$, which yields $A=1$.
I know the final answer is this:
$$\frac{1}{x+2} - \frac{x-1}{x^2+x+1} - \frac{1}{(x^2+x+1)^2}$$
But I've worked this problem many ways and cannot make progress on the numerators for the other fractions.
 A: $$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}= \frac {(3x^2+3x +3)-(x+2)}{(x+2)(x^2+x+1)^2}=$$
$$ \frac {3}{(x+2)(x^2+x+1)}-\frac { 1}{(x^2+x+1)^2} $$
Now you may proceed with the first fraction. 
$$ \frac {3}{(x+2)(x^2+x+1)}=\frac {1}{x+2} -\frac {x-1}{x^2+x+1}$$
A: I'll expand all the things from the comment.
Method 1:
By expanding we get
$$
\begin{cases}
A+B=0&&\hbox{ for }x^4\\\
2A+3B+C=0&&\hbox{ for }x^3\\\
3 A + 3 B + 3 C + D - 3=0&&\hbox{ for }x^2\\\
2 A + 2 B + 3 C + 2 D + E - 2=0&&\hbox{ for }x\\\
A + 2 C + 2 E - 1=0&&\hbox{ for }1
\end{cases}
$$
which boils down to $A = 1, B = -1, C = 1, D = 0, E = -1$.
Method 2 "substitution":
with $x=-\frac12\pm\frac{\sqrt{3}}{2}i$ we get rid of $(x^2+x+1)$, can have equations for $D,E$ and after finding $D,E$ and bringing them to the LHS we can cancel by $(x+2)$ which boils down to Mohammad Riazi-Kermani's answer
A: Now that you know $A=1$ set $A$ equal to $1$. You get
$$3x^2+2x+1 = (x^2+x+1)^2 + (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$3x^2+2x+1 - (x^4+2x^3+3x^2+2x+1) = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$-x^4-2x^3 = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$-x^3(x+2) = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$-x^3 = (Bx+C)(x^2+x+1) + (Dx+E)$$
$$-x^3 = Bx^3 + (B+C)x^2 + (B+C+D)x + (C+E)$$
and proceed from there
