Explanation:
(A) : We have,
\(\int \frac{9 x+15}{x^{3}-6 x-9} d x=A \log |g(x)|+B \log |f(x)|+C\)
Now, \(\frac{9 x+15}{x^{3}-6 x-9}=\frac{9 x+15}{(x-3)\left(x^{2}+3 x+3\right)}\)
\(\therefore \frac{9 \mathrm{x}+15}{(\mathrm{x}-3)\left(\mathrm{x}^{2}+3 \mathrm{x}+3\right)}=\frac{\mathrm{A}}{\mathrm{x}-3}+\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+3 \mathrm{x}+3}\)
\(\Rightarrow 9 \mathrm{x}+15=\mathrm{A}\left(\mathrm{x}^{2}+3 \mathrm{x}+3\right)+(\mathrm{Bx}+\mathrm{C})(\mathrm{x}-3)\)
\(\Rightarrow 9 x+15=x^{2}(A+B)+x(3 A-3 B+C)+(3 A-3 C)\)
On equating coefficients of \(x^{2}, x\) and constant term, we get
\(\mathrm{A}+\mathrm{B}=0,3 \mathrm{~A}-3 \mathrm{~B}+\mathrm{C}=9\) and \(3 \mathrm{~A}-3 \mathrm{C}=15\)
On solving above three equations
We get, \(\mathrm{A}=2, \mathrm{~B}=-2\) and \(\mathrm{C}=-3\)
\(\therefore \int \frac{9 x+15}{(x-3)\left(x^{2}+3 x+3\right)} d x\)
\(\quad \int \frac{2}{x-3} d x+\int \frac{-2 x-3}{x^{2}+3 x+3} d x\)
\(\quad \int \frac{2}{x-3} d x-\int \frac{2 x+3}{x^{2}+3 x+3} d x\)
\(\quad 2 \log |x-3|-\log \left|x^{2}+3 x+3\right|+C\)
\(\therefore \mathrm{A}=2, \mathrm{~B}=-1, \mathrm{~g}(\mathrm{x})=\mathrm{x}-3\) and \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+3 \mathrm{x}+3\)
\(\therefore \frac{(\mathrm{A}-\mathrm{B}) \mathrm{g}(4)}{\mathrm{f}(-1)}=\frac{(2+1)(4-3)}{(-1)^{2}+3(-1)+3}=3\)