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

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(D) Given the partial fraction decomposition: $\frac{x^2-2}{(x^2+1)(x^2+3)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}$.Let $y = x^2$. The expression b

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$\frac{5}{2}$

Vedclass · Answer

(D) Given the partial fraction decomposition: $\frac{x^2-2}{(x^2+1)(x^2+3)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}$.Let $y = x^2$. The expression becomes $\frac{y-2}{(y+1)(y+3)} = \frac{B}{y+1} + \frac{D}{y+3}$ (since the numerator terms $Ax$ and $Cx$ must be $0$ for the equality to hold for all $x$).To find $D$, we use the cover-up method by multiplying both sides by $(y+3)$ and setting $y = -3$:$D = \left[ \frac{y-2}{y+1} \right]_{y=-3} = \frac{-3-2}{-3+1} = \frac{-5}{-2} = \frac{5}{2}$.

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

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