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

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(NONE) Given $\frac{x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$.Multiplying both sides by $(x-1)^2(x^2+1)$,we get $

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(NONE) Given $\frac{x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$.Multiplying both sides by $(x-1)^2(x^2+1)$,we get $x+1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2$.For $x=1$,$1+1 = B(1^2+1) \implies 2 = 2B \implies B=1$.Comparing coefficients of $x^3$: $0 = A+C \implies C = -A$.Comparing coefficients of $x^0$: $1 = -A + B + D \implies 1 = -A + 1 + D \implies D = A$.Comparing coefficients of $x^2$: $0 = -A + B + C - 2D \implies 0 = -A + 1 - A - 2A \implies 4A = 1 \implies A = \frac{1}{4}$.Thus,$A = \frac{1}{4}$,$B = 1$,$C = -\frac{1}{4}$,$D = \frac{1}{4}$.Now,$\sqrt{3A^2+4D^2+5C^2+B^2} = \sqrt{3(\frac{1}{16}) + 4(\frac{1}{16}) + 5(\frac{1}{16}) + 1} = \sqrt{\frac{3+4+5}{16} + 1} = \sqrt{\frac{12}{16} + 1} = \sqrt{\frac{3}{4} + 1} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}$.

If $\frac{x+1}{(x-1)^2(x^2+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}$,then $\sqrt{3A^2+4D^2+5C^2+B^2}=$

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