If frac{x^2}{(x^2+2)(x^4-1)} = frac{A}{x^2-1} | vedclass

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(B) Let $y = x^2$. The expression becomes $\frac{y}{(y+2)(y^2-1)} = \frac{y}{(y+2)(y-1)(y+1)} = \frac{A}{y-1} + \frac{B}{y+1} + \frac{C}{y+2}$.Using p

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$\frac{4}{3}$

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(B) Let $y = x^2$. The expression becomes $\frac{y}{(y+2)(y^2-1)} = \frac{y}{(y+2)(y-1)(y+1)} = \frac{A}{y-1} + \frac{B}{y+1} + \frac{C}{y+2}$.Using partial fraction decomposition:$y = A(y+1)(y+2) + B(y-1)(y+2) + C(y-1)(y+1)$.For $y=1$: $1 = A(2)(3) \implies 6A = 1 \implies A = \frac{1}{6}$.For $y=-1$: $-1 = B(-2)(1) \implies -2B = -1 \implies B = \frac{1}{2}$.For $y=-2$: $-2 = C(-3)(-1) \implies 3C = -2 \implies C = -\frac{2}{3}$.Thus,$A+B-C = \frac{1}{6} + \frac{1}{2} - (-\frac{2}{3}) = \frac{1}{6} + \frac{3}{6} + \frac{4}{6} = \frac{8}{6} = \frac{4}{3}$.

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

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