Split into partial fractions $\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}$

Question

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.

Answer 1

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

Answer 2

$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}= \frac {(3x^2+3x +3)-(x+2)}{(x+2)(x^2+x+1)^2}=$$

$$ \frac {3}{(x+2)(x^2+x+1)}-\frac { 1}{(x^2+x+1)^2} $$

Now you may proceed with the first fraction.

$$ \frac {3}{(x+2)(x^2+x+1)}=\frac {1}{x+2} -\frac {x-1}{x^2+x+1}$$

Answer 3

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