begin{aligned} & frac{x^2+1}{x^4+4}=frac{A x+ | vedclass

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(C) Given: $\frac{x^2+1}{x^4+4}=\frac{A x+B}{x^2-2 x+2}+\frac{C x+D}{x^2+2 x+2}$ Expanding the right side: $\frac{x^2+1}{x^4+4} = \frac{(A x+B)(x^2+2

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$2 D$

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(C) Given: $\frac{x^2+1}{x^4+4}=\frac{A x+B}{x^2-2 x+2}+\frac{C x+D}{x^2+2 x+2}$ Expanding the right side: $\frac{x^2+1}{x^4+4} = \frac{(A x+B)(x^2+2 x+2)+(C x+D)(x^2-2 x+2)}{(x^2-2 x+2)(x^2+2 x+2)}$ Note that $(x^2-2 x+2)(x^2+2 x+2) = ((x^2+2)-2x)((x^2+2)+2x) = (x^2+2)^2 - (2x)^2 = x^4+4x^2+4-4x^2 = x^4+4$. Equating the numerators: $x^2+1 = (A+C)x^3 + (2A+B-2C+D)x^2 + (2A+2B+2C-2D)x + (2B+2D)$. Comparing coefficients: $1) A+C=0 \Rightarrow A=-C$ $2) 2A+B-2C+D=1$ $3) 2A+2B+2C-2D=0 \Rightarrow A+B+C-D=0$ $4) 2B+2D=1$ From $(1)$,$A+C=0$,so $(3)$ becomes $B-D=0 \Rightarrow B=D$. Substituting $B=D$ into $(4)$,$2B+2B=1 \Rightarrow 4B=1 \Rightarrow B=D=\frac{1}{4}$. From $(2)$,$2(A-C)+B+D=1$. Since $A=-C$,$2(2A)+2B=1 \Rightarrow 4A+2B=1$. Substituting $B=\frac{1}{4}$,$4A + \frac{1}{2} = 1 \Rightarrow 4A = \frac{1}{2} \Rightarrow A = \frac{1}{8}$. Thus $C = -\frac{1}{8}$. Finally,$3A+2B+3C = 3(A+C) + 2B = 3(0) + 2(\frac{1}{4}) = \frac{1}{2} = 2D$.

$\begin{aligned} & \frac{x^2+1}{x^4+4}=\frac{A x+B}{x^2-2 x+2}+\frac{C x+D}{x^2+2 x+2} \\ & \Rightarrow 3 A+2 B+3 C=\end{aligned}$

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