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

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

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$\frac{1}{2}$

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(B) Given the partial fraction decomposition: $\frac{x+1}{\left(x^2+1\right)(x-1)^2}=\frac{A x+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}$.Multiplying both sides by $(x^2+1)(x-1)^2$, we get:$x+1 = (Ax+B)(x-1)^2 + C(x-1)(x^2+1) + D(x^2+1)$.Setting $x=1$: $1+1 = D(1^2+1) \Rightarrow 2 = 2D \Rightarrow D=1$.Expanding the right side:$x+1 = (Ax+B)(x^2-2x+1) + C(x^3-x^2+x-1) + D(x^2+1)$$x+1 = A(x^3-2x^2+x) + B(x^2-2x+1) + C(x^3-x^2+x-1) + D(x^2+1)$$x+1 = (A+C)x^3 + (-2A+B-C+D)x^2 + (A-2B+C)x + (B-C+D)$.Comparing coefficients:$1$) $A+C = 0 \Rightarrow C = -A$$2$) $-2A+B-C+D = 0 \Rightarrow -2A+B-(-A)+1 = 0 \Rightarrow -A+B+1 = 0 \Rightarrow B = A-1$$3$) $A-2B+C = 1 \Rightarrow A-2(A-1)+(-A) = 1 \Rightarrow A-2A+2-A = 1 \Rightarrow -2A = -1 \Rightarrow A = \frac{1}{2}$.Then $C = -\frac{1}{2}$ and $B = \frac{1}{2}-1 = -\frac{1}{2}$.Finally, $A+B+C+D = \frac{1}{2} - \frac{1}{2} - \frac{1}{2} + 1 = \frac{1}{2}$.

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

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