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

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

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$-\frac{5}{2}$

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(C) Given the partial fraction decomposition: $\frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}$ Multiplying both sides by $(x+1)^2(x+3)$,we get: $3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$ Setting $x = -1$: $3(-1)-2 = B(-1+3)$ $\Rightarrow -5 = 2B$ $\Rightarrow B = -\frac{5}{2}$ Setting $x = -3$: $3(-3)-2 = C(-3+1)^2$ $\Rightarrow -11 = 4C$ $\Rightarrow C = -\frac{11}{4}$ Comparing the coefficients of $x^2$ on both sides: $0 = A + C \Rightarrow A = -C = \frac{11}{4}$ Thus,$A+B+C = \frac{11}{4} - \frac{5}{2} - \frac{11}{4} = -\frac{5}{2}$

If $\frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}$,then $A+B+C=$

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