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

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(A) Given the partial fraction decomposition: $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$ Equating the numerators: $3x+2 = A(2x

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$2$

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(A) Given the partial fraction decomposition: $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$ Equating the numerators: $3x+2 = A(2x^2+3) + (x+1)(Bx+C)$ $3x+2 = 2Ax^2 + 3A + Bx^2 + Cx + Bx + C$ $3x+2 = (2A+B)x^2 + (B+C)x + (3A+C)$ Comparing coefficients: $2A+B = 0$ $(i)$ $B+C = 3$ (ii) $3A+C = 2$ (iii) From $(i)$,$B = -2A$. Substitute into (ii): $-2A+C = 3$ (iv) Subtract (iv) from (iii): $(3A+C) - (-2A+C) = 2 - 3 \implies 5A = -1 \implies A = -\frac{1}{5}$ Then $B = -2(-\frac{1}{5}) = \frac{2}{5}$ And $C = 3 - B = 3 - \frac{2}{5} = \frac{13}{5}$ Finally,$A-B+C = -\frac{1}{5} - \frac{2}{5} + \frac{13}{5} = \frac{10}{5} = 2$

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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