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

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

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$2$

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(B) Given the partial fraction decomposition: $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$.Multiplying both sides by $(x+1)(2x^2+3)$,we get: $3x+2 = A(2x^2+3) + (x+1)(Bx+C)$.Setting $x = -1$: $3(-1)+2 = A(2(-1)^2+3) + 0$ $\Rightarrow -1 = 5A$ $\Rightarrow A = -\frac{1}{5}$.Comparing the coefficients of $x^2$: $0 = 2A + B \Rightarrow B = -2A = -2(-\frac{1}{5}) = \frac{2}{5}$.Comparing the constant terms: $2 = 3A + C \Rightarrow C = 2 - 3A = 2 - 3(-\frac{1}{5}) = 2 + \frac{3}{5} = \frac{13}{5}$.Now,calculate $A-B+C = -\frac{1}{5} - \frac{2}{5} + \frac{13}{5} = \frac{-1-2+13}{5} = \frac{10}{5} = 2$.

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

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