If fleft(frac{2 x+3}{3 x+5}right)=x+4, where | vedclass

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(D) Given $f\left(\frac{2 x+3}{3 x+5}\right)=x+4$. Let $y = \frac{2 x+3}{3 x+5}$.Then $y(3x+5) = 2x+3 \Rightarrow 3xy + 5y = 2x+3 \Rightarrow x(3y-2)

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$\frac{-8}{3}$

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(D) Given $f\left(\frac{2 x+3}{3 x+5}\right)=x+4$. Let $y = \frac{2 x+3}{3 x+5}$.Then $y(3x+5) = 2x+3 \Rightarrow 3xy + 5y = 2x+3 \Rightarrow x(3y-2) = 3-5y \Rightarrow x = \frac{3-5y}{3y-2}$.Substituting $x$ in $f(y) = x+4$, we get $f(y) = \frac{3-5y}{3y-2} + 4 = \frac{3-5y + 12y - 8}{3y-2} = \frac{7y-5}{3y-2}$.Thus, $f(x) = \frac{7x-5}{3x-2}$.Now, $\int f(x) dx = \int \frac{7x-5}{3x-2} dx = \int \frac{\frac{7}{3}(3x-2) - 5 + \frac{14}{3}}{3x-2} dx = \int \left( \frac{7}{3} - \frac{1/3}{3x-2} \right) dx$.$= \frac{7}{3}x - \frac{1}{9} \ln |3x-2| + C$.Comparing with $Ax + B \ln |3x-2| + C$, we get $A = \frac{7}{3}$ and $B = -\frac{1}{9}$.Therefore, $3B - A = 3(-\frac{1}{9}) - \frac{7}{3} = -\frac{1}{3} - \frac{7}{3} = -\frac{8}{3}$.

If $f\left(\frac{2 x+3}{3 x+5}\right)=x+4$, where $x \neq \frac{-5}{3}, \frac{-2}{3}$, and $\int f(x) d x=A x+B \ln |3 x-2|+C$, then $3 B-A=$

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