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

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Question

(A) Given the partial fraction decomposition: $\frac{(x - 1)^2}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$.Expanding the numerator on the lef

Vedclass · Accepted Answer

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

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(A) Given the partial fraction decomposition: $\frac{(x - 1)^2}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$.Expanding the numerator on the left side: $\frac{x^2 - 2x + 1}{x(x^2 + 1)} = \frac{A(x^2 + 1) + x(Bx + C)}{x(x^2 + 1)}$.Equating the numerators: $x^2 - 2x + 1 = A(x^2 + 1) + Bx^2 + Cx$.$x^2 - 2x + 1 = (A + B)x^2 + Cx + A$.Comparing the coefficients of $x^2$,$x$,and the constant term:$1$) $A + B = 1$$2$) $C = -2$$3$) $A = 1$Substituting $A = 1$ into equation $(1)$: $1 + B = 1 \implies B = 0$.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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