The partial fraction decomposition of frac{9x | vedclass

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(D) The partial fraction form for $\frac{px+q}{(x+a)(x^2+b^2)}$ is $\frac{A}{x+a}+\frac{Bx+C}{x^2+b^2}$.Set $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3}

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$\frac{-17}{5(x+3)}+\frac{(17x-6)}{5(x^2+1)}$

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(D) The partial fraction form for $\frac{px+q}{(x+a)(x^2+b^2)}$ is $\frac{A}{x+a}+\frac{Bx+C}{x^2+b^2}$.Set $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$.Multiplying both sides by $(x+3)(x^2+1)$, we get $9x-7 = A(x^2+1) + (Bx+C)(x+3)$.Expanding the right side: $9x-7 = Ax^2 + A + Bx^2 + 3Bx + Cx + 3C = (A+B)x^2 + (3B+C)x + (A+3C)$.Comparing coefficients:$A+B = 0 \implies B = -A$$3B+C = 9$$A+3C = -7$Substituting $B = -A$ into the second equation: $3(-A) + C = 9 \implies -3A + C = 9 \implies C = 9 + 3A$.Substituting $C$ into the third equation: $A + 3(9 + 3A) = -7 \implies A + 27 + 9A = -7 \implies 10A = -34 \implies A = -\frac{34}{10} = -\frac{17}{5}$.Then $B = -A = \frac{17}{5}$ and $C = 9 + 3(-\frac{17}{5}) = \frac{45-51}{5} = -\frac{6}{5}$.Substituting these values back, we get $\frac{-17}{5(x+3)} + \frac{\frac{17}{5}x - \frac{6}{5}}{x^2+1} = \frac{-17}{5(x+3)} + \frac{17x-6}{5(x^2+1)}$.

The partial fraction decomposition of $\frac{9x-7}{(x+3)(x^2+1)}$ is

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