If frac{3x+1}{(x-1)^2(x^2+1)} = frac{A}{x-1} | vedclass

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

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(D) Given the partial fraction decomposition: $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$.Multiplying both sides by $(x-1)^2(x^2+1)$, we get: $3x+1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2$.Setting $x=1$: $3(1)+1 = B(1^2+1) \implies 4 = 2B \implies B=2$.Expanding the right side: $3x+1 = A(x^3-x^2+x-1) + 2(x^2+1) + (Cx+D)(x^2-2x+1)$.Comparing coefficients of $x^3$: $0 = A+C \implies C = -A$.Comparing coefficients of $x^2$: $0 = -A+2+D-2C = -A+2+D+2A = A+D+2 \implies D = -A-2$.Comparing constant terms: $1 = -A+2+D \implies -A+D = -1$.Substituting $D = -A-2$: $-A-A-2 = -1 \implies -2A = 1 \implies A = -1/2$.Then $C = 1/2$ and $D = -(-1/2)-2 = 1/2-2 = -3/2$.We need to calculate $2(A-C+B+D) = 2(-1/2 - 1/2 + 2 - 3/2) = 2(-1 + 2 - 1.5) = 2(-0.5) = -1$.

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

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