frac{2x^2+1}{x^3-1} = frac{A}{x-1} + frac{Bx+ | vedclass

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(B) Given the partial fraction decomposition: $\frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1}$.Since $x^3-1 = (x-1)(x^2+x+1)$,we have $2x

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Given the partial fraction decomposition: $\frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1}$.Since $x^3-1 = (x-1)(x^2+x+1)$,we have $2x^2+1 = A(x^2+x+1) + (Bx+C)(x-1)$.Expanding the right side: $2x^2+1 = Ax^2 + Ax + A + Bx^2 - Bx + Cx - C$.Grouping terms by powers of $x$: $2x^2 + 0x + 1 = (A+B)x^2 + (A-B+C)x + (A-C)$.Comparing coefficients:$1) A+B = 2$$2) A-B+C = 0$$3) A-C = 1$From $(3)$,$C = A-1$. Substituting into $(2)$: $A-B+(A-1) = 0 \Rightarrow 2A-B = 1$.Adding this to $(1)$: $(2A-B) + (A+B) = 1+2$ $\Rightarrow 3A = 3$ $\Rightarrow A = 1$.Then $B = 2-A = 1$ and $C = A-1 = 0$.Finally,$7A + 2B + C = 7(1) + 2(1) + 0 = 9$.

$\frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \Rightarrow 7A + 2B + C = ?$

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