If frac{-x^2+6x+1}{(x-1)^2(x^2+2)} = frac{A}{ | vedclass

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(D) Given the partial fraction decomposition: $\frac{-x^2+6x+1}{(x-1)^2(x^2+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx-3}{x^2+2}$.Multiplying

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(D) Given the partial fraction decomposition: $\frac{-x^2+6x+1}{(x-1)^2(x^2+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx-3}{x^2+2}$.Multiplying both sides by $(x-1)^2(x^2+2)$,we get:$-x^2+6x+1 = A(x-1)(x^2+2) + B(x^2+2) + (Cx-3)(x-1)^2$.Setting $x=1$: $-1+6+1 = B(1+2) \Rightarrow 6 = 3B \Rightarrow B=2$.Expanding the right side: $-x^2+6x+1 = A(x^3-x^2+2x-2) + B(x^2+2) + (Cx-3)(x^2-2x+1)$.Comparing the coefficients of $x^3$: $0 = A + C \Rightarrow C = -A$.Comparing the coefficients of $x^2$: $-1 = -A + B + (-2C - 3) = -A + 2 - 2C - 3 = -A - 2C - 1$.Thus,$A + 2C = 0$. Since $C = -A$,we have $A - 2A = 0 \Rightarrow -A = 0 \Rightarrow A = 0$.Therefore,$C = -A = 0$.Finally,$A+B+C = 0 + 2 + 0 = 2$.

If $\frac{-x^2+6x+1}{(x-1)^2(x^2+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx-3}{x^2+2}$,then $A+B+C=$

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