If frac{2x+1}{(x-1)^2(x^2+1)}=frac{A}{x-1}+fr | vedclass

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

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$\frac{1}{2}$

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

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

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