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

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(B) Given the partial fraction decomposition: $\frac{(x - 1)^2}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$.Multiplying both sides by $x(x^2 +

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$A = 1, B = 0, C = -2$

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(B) Given the partial fraction decomposition: $\frac{(x - 1)^2}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$.Multiplying both sides by $x(x^2 + 1)$,we get: $(x - 1)^2 = A(x^2 + 1) + x(Bx + C)$.Expanding the left side: $x^2 - 2x + 1 = Ax^2 + A + Bx^2 + Cx$.Grouping terms by powers of $x$: $x^2 - 2x + 1 = (A + B)x^2 + Cx + A$.Comparing coefficients on both sides:$1$) Constant term: $A = 1$.$2$) Coefficient of $x$: $C = -2$.$3$) Coefficient of $x^2$: $A + B = 0$. Since $A = 1$,then $1 + B = 0$,which gives $B = -1$.Wait,let us re-evaluate: $A + B = 1$. Since $A = 1$,then $1 + B = 1$,which gives $B = 0$.Thus,$A = 1, B = 0, C = -2$.

If $\frac{(x - 1)^2}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$,then

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