If int frac{9x+15}{x^3-6x-9} dx = A log |g(x) | vedclass

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(A) We have, $\int \frac{9x+15}{x^3-6x-9} dx = A \log |g(x)| + B \log |f(x)| + C$. First, factor the denominator: $x^3-6x-9 = (x-3)(x^2+3x+3)$. Using

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$3$

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(A) We have, $\int \frac{9x+15}{x^3-6x-9} dx = A \log |g(x)| + B \log |f(x)| + C$. First, factor the denominator: $x^3-6x-9 = (x-3)(x^2+3x+3)$. Using partial fractions: $\frac{9x+15}{(x-3)(x^2+3x+3)} = \frac{A}{x-3} + \frac{Bx+D}{x^2+3x+3}$. $9x+15 = A(x^2+3x+3) + (Bx+D)(x-3)$. For $x=3$: $9(3)+15 = A(9+9+3) \Rightarrow 42 = 21A \Rightarrow A=2$. Comparing coefficients of $x^2$: $A+B=0 \Rightarrow B=-2$. Comparing constant terms: $3A-3D=15 \Rightarrow 3(2)-3D=15 \Rightarrow 6-3D=15 \Rightarrow -3D=9 \Rightarrow D=-3$. Thus, $\int \frac{2}{x-3} dx - \int \frac{2x+3}{x^2+3x+3} dx = 2 \log |x-3| - \log |x^2+3x+3| + C$. Comparing with $A \log |g(x)| + B \log |f(x)| + C$, we get $A=2, g(x)=x-3, B=-1, f(x)=x^2+3x+3$. Then, $\frac{(A-B)g(4)}{f(-1)} = \frac{(2 - (-1))(4-3)}{(-1)^2 + 3(-1) + 3} = \frac{3(1)}{1-3+3} = \frac{3}{1} = 3$.

If $\int \frac{9x+15}{x^3-6x-9} dx = A \log |g(x)| + B \log |f(x)| + C$, then $\frac{(A-B) g(4)}{f(-1)} =$

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