If frac{x^2+5}{(x^2+1)(x-2)}=frac{A}{x-2}+fra | vedclass

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(C) Given the partial fraction decomposition: $\frac{x^2+5}{(x^2+1)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$.Multiplying both sides by $(x^2+1)(x-2)$,

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$-3/5$

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(C) Given the partial fraction decomposition: $\frac{x^2+5}{(x^2+1)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$.Multiplying both sides by $(x^2+1)(x-2)$, we get: $x^2+5 = A(x^2+1) + (Bx+C)(x-2)$.Expanding the right side: $x^2+5 = Ax^2 + A + Bx^2 - 2Bx + Cx - 2C$.Grouping terms by powers of $x$: $x^2+5 = (A+B)x^2 + (C-2B)x + (A-2C)$.Comparing coefficients on both sides:$1$) $A+B = 1$$2$) $C-2B = 0 \Rightarrow C = 2B$$3$) $A-2C = 5$Substitute $C=2B$ into equation $(3)$: $A - 2(2B) = 5 \Rightarrow A - 4B = 5$.Now solve the system of equations:$A+B = 1$$A-4B = 5$Subtracting the second from the first: $5B = -4 \Rightarrow B = -\frac{4}{5}$.Then $A = 1 - B = 1 - (-4/5) = 9/5$.And $C = 2B = 2(-4/5) = -8/5$.Finally, $A+B+C = \frac{9}{5} - \frac{4}{5} - \frac{8}{5} = \frac{9-4-8}{5} = -\frac{3}{5}$.

If $\frac{x^2+5}{(x^2+1)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$, then $A+B+C=$

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