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

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

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$\frac{19}{5}$

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(D) Given the partial fraction decomposition: $\frac{x^2-3}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1}$.Multiplying both sides by $(x+2)(x^2+1)$, we get: $x^2-3 = A(x^2+1) + (Bx+C)(x+2)$.To find $A$, set $x = -2$: $(-2)^2 - 3 = A((-2)^2 + 1) \implies 4-3 = A(4+1) \implies 1 = 5A \implies A = \frac{1}{5}$.Expanding the right side: $x^2-3 = A x^2 + A + B x^2 + 2Bx + Cx + 2C = (A+B)x^2 + (2B+C)x + (A+2C)$.Comparing coefficients of $x^2$: $A+B = 1 \implies \frac{1}{5} + B = 1 \implies B = \frac{4}{5}$.Comparing coefficients of $x$: $2B+C = 0 \implies 2(\frac{4}{5}) + C = 0 \implies C = -\frac{8}{5}$.Now calculate $3A+2B-C = 3(\frac{1}{5}) + 2(\frac{4}{5}) - (-\frac{8}{5}) = \frac{3}{5} + \frac{8}{5} + \frac{8}{5} = \frac{19}{5}$.

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

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