If frac{x^2+3}{x^4+2 x^2+9}=frac{A x+B}{x^2+a | vedclass

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(D) We factor the denominator: $x^4+2 x^2+9 = (x^4+6 x^2+9) - 4 x^2 = (x^2+3)^2 - (2 x)^2 = (x^2-2 x+3)(x^2+2 x+3)$.Using partial fractions,we have $\

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$2$

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(D) We factor the denominator: $x^4+2 x^2+9 = (x^4+6 x^2+9) - 4 x^2 = (x^2+3)^2 - (2 x)^2 = (x^2-2 x+3)(x^2+2 x+3)$.Using partial fractions,we have $\frac{x^2+3}{(x^2-2 x+3)(x^2+2 x+3)} = \frac{Ax+B}{x^2-2 x+3} + \frac{Cx+D}{x^2+2 x+3}$.By solving,we get $\frac{x^2+3}{x^4+2 x^2+9} = \frac{1}{2} \left( \frac{1}{x^2-2 x+3} + \frac{1}{x^2+2 x+3} \right) = \frac{0x + 1/2}{x^2-2 x+3} + \frac{0x + 1/2}{x^2+2 x+3}$.Comparing coefficients,we identify $A=0, B=1/2, a=-2, b=3, C=0, D=1/2, c=2$.Substituting these values into the expression $aA+bB+cC+D$:$(-2)(0) + (3)(1/2) + (2)(0) + 1/2 = 0 + 3/2 + 0 + 1/2 = 4/2 = 2$.

If $\frac{x^2+3}{x^4+2 x^2+9}=\frac{A x+B}{x^2+a x+b}+\frac{C x+D}{x^2+c x+b}$ then $a A+b B+c C+D=$

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