If frac{x+1}{x^3(x-1)} = frac{a}{x} + frac{b} | vedclass

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(B) Given the partial fraction decomposition: $\frac{x+1}{x^3(x-1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{d}{x-1}$.Multiplying both si

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$a = b = 2c = -d$

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(B) Given the partial fraction decomposition: $\frac{x+1}{x^3(x-1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{d}{x-1}$.Multiplying both sides by $x^3(x-1)$,we get: $x+1 = ax^2(x-1) + bx(x-1) + c(x-1) + dx^3$.Expanding the right side: $x+1 = a(x^3 - x^2) + b(x^2 - x) + c(x-1) + dx^3$.Grouping the powers of $x$: $x+1 = (a+d)x^3 + (b-a)x^2 + (c-b)x - c$.Comparing coefficients:Constant term: $-c = 1 \implies c = -1$.Coefficient of $x$: $c - b = 1 \implies -1 - b = 1 \implies b = -2$.Coefficient of $x^2$: $b - a = 0 \implies a = b = -2$.Coefficient of $x^3$: $a + d = 0 \implies d = -a = 2$.Checking the values: $a = -2, b = -2, c = -1, d = 2$.We observe that $a = b = 2c = -d$ since $-2 = -2 = 2(-1) = -(2)$.

If $\frac{x+1}{x^3(x-1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{d}{x-1}$,then:

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