If frac{9x-7}{(x+3)(x^2+1)} = frac{A}{x+3} + | vedclass

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(B) Given, $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$ Equating the numerators: $9x-7 = A(x^2+1) + (Bx+C)(x+3)$ $9x-7 = Ax^2 + A

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$\frac{-6}{5}$

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(B) Given, $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$ Equating the numerators: $9x-7 = A(x^2+1) + (Bx+C)(x+3)$ $9x-7 = Ax^2 + A + Bx^2 + 3Bx + Cx + 3C$ $9x-7 = (A+B)x^2 + (3B+C)x + (A+3C)$ Comparing coefficients on both sides: $A+B = 0$ $3B+C = 9$ $A+3C = -7$ From $A+B=0$, we get $B = -A$. Substituting $B = -A$ into $3B+C=9$, we get $-3A+C=9$, so $C = 9+3A$. Substituting $C = 9+3A$ into $A+3C=-7$: $A + 3(9+3A) = -7$ $A + 27 + 9A = -7$ $10A = -34 \implies A = -\frac{34}{10} = -\frac{17}{5}$ Then $B = -A = \frac{17}{5}$ And $C = 9 + 3(-\frac{17}{5}) = 9 - \frac{51}{5} = \frac{45-51}{5} = -\frac{6}{5}$ Finally, $A+B+C = -\frac{17}{5} + \frac{17}{5} - \frac{6}{5} = -\frac{6}{5}$

If $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$, where $A, B, C \in \mathbb{R}$, then $A+B+C = $

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