The partial fraction decomposition of frac{3x | vedclass

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(C) We express the fraction as: $\frac{3x+1}{(x+2)(x-1)^2} = \frac{A}{x+2} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$Multiplying both sides by $(x+2)(x-1)^2

Vedclass · Accepted Answer

$\frac{-5}{9} \left(\frac{1}{x+2}\right) + \frac{5}{9} \cdot \frac{1}{x-1} + \frac{4}{3} \cdot \frac{1}{(x-1)^2}$

Vedclass · Answer

(C) We express the fraction as: $\frac{3x+1}{(x+2)(x-1)^2} = \frac{A}{x+2} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$Multiplying both sides by $(x+2)(x-1)^2$,we get:$3x+1 = A(x-1)^2 + B(x-1)(x+2) + C(x+2)$Expanding the terms:$3x+1 = A(x^2 - 2x + 1) + B(x^2 + x - 2) + C(x+2)$$3x+1 = (A+B)x^2 + (-2A + B + C)x + (A - 2B + 2C)$Comparing coefficients:$1$) $A+B = 0 \Rightarrow A = -B$$2$) $-2A + B + C = 3$$3$) $A - 2B + 2C = 1$Substituting $A = -B$ into $(2)$ and $(3)$:$2$) $2B + B + C = 3 \Rightarrow 3B + C = 3$$3$) $-B - 2B + 2C = 1 \Rightarrow -3B + 2C = 1$Adding the two equations: $(3B + C) + (-3B + 2C) = 3 + 1$ $\Rightarrow 3C = 4$ $\Rightarrow C = \frac{4}{3}$Substituting $C = \frac{4}{3}$ into $3B + C = 3$:$3B + \frac{4}{3} = 3$ $\Rightarrow 3B = \frac{5}{3}$ $\Rightarrow B = \frac{5}{9}$Since $A = -B$,$A = -\frac{5}{9}$Thus,the decomposition is $\frac{-5}{9(x+2)} + \frac{5}{9(x-1)} + \frac{4}{3(x-1)^2}$.

The partial fraction decomposition of $\frac{3x+1}{(x-1)^2(x+2)}$ is:

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The values of $x$ for which $\frac{x}{(x-1)^2(x-2)}$ has a power series expansion and the coefficient of $x^n$ in such expansion are respectively:

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