If frac{x^2-x+1}{(x^2+1)(x^2+x+1)}=frac{Ax+B} | vedclass

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(D) Given the partial fraction decomposition: $\frac{x^2-x+1}{(x^2+1)(x^2+x+1)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+x+1}$ Multiplying both sides by $(x

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$2$

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(D) Given the partial fraction decomposition: $\frac{x^2-x+1}{(x^2+1)(x^2+x+1)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+x+1}$ Multiplying both sides by $(x^2+1)(x^2+x+1)$, we get: $x^2-x+1 = (Ax+B)(x^2+x+1) + (Cx+D)(x^2+1)$ $x^2-x+1 = Ax^3 + Ax^2 + Ax + Bx^2 + Bx + B + Cx^3 + Cx + Dx^2 + D$ $x^2-x+1 = (A+C)x^3 + (A+B+D)x^2 + (A+B+C)x + (B+D)$ Comparing coefficients on both sides: $1) A+C = 0$ $2) A+B+D = 1$ $3) A+B+C = -1$ $4) B+D = 1$ From $(1)$, $C = -A$. Substituting into $(3)$: $A+B-A = -1 \Rightarrow B = -1$. Using $B = -1$ in $(4)$: $-1+D = 1 \Rightarrow D = 2$. Using $B = -1$ and $D = 2$ in $(2)$: $A-1+2 = 1 \Rightarrow A = 0$. Then $C = -A = 0$. Thus, $A=0, B=-1, C=0, D=2$. Calculating $A+2B+C+2D = 0 + 2(-1) + 0 + 2(2) = -2 + 4 = 2$.

If $\frac{x^2-x+1}{(x^2+1)(x^2+x+1)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+x+1}$, then $A+2B+C+2D=$

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