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

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

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$(1, -1, 0)$

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(A) Given the partial fraction decomposition: $\frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$ Multiply both sides by $x(x^2 + 1)$: $1 = A(x^2 + 1) + (Bx + C)x$ $1 = Ax^2 + A + Bx^2 + Cx$ $1 = (A + B)x^2 + Cx + A$ Comparing the coefficients of $x^2$,$x$,and the constant term on both sides: $A + B = 0$ $C = 0$ $A = 1$ Substituting $A = 1$ into $A + B = 0$,we get $1 + B = 0$,which implies $B = -1$. Thus,$(A, B, C) = (1, -1, 0)$.

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

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