<s>[INST] 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.
 [/INST] 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
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