We have, $f\left(\frac{2 x+3}{3 x+5}\right)=x+4$
$\begin{aligned}
& \text { Put, } \frac{2 x+3}{3 x+5}=y \Rightarrow 2 x+3=3 x y+5 y \Rightarrow x=\frac{5 y-3}{2-3 y} \\
& \therefore f(y)=\frac{5 y-3}{2-3 y}+4=\frac{5 y-3+8-12 y}{2-3 y} \Rightarrow f(y)=\frac{7 y-5}{3 x-2} \\
& \therefore \quad f(x)=\frac{7 x-5}{3 x-2} \Rightarrow \int f(x) d x=\int \frac{7 x-5}{3 x-2} d x
\end{aligned}$
Put, $3 x-2=t \Rightarrow x=\frac{t+2}{3}$
$\begin{aligned}
d x & =\frac{d t}{3}=\frac{1}{3} \int \frac{7\left(\frac{t+2}{3}\right)-5}{t} d t \\
& =\frac{1}{9} \int \frac{7 t-1}{t} d t=\frac{7}{9} \int d t-\frac{1}{3} \int \frac{d t}{t} \\
& =\frac{7}{9} t-\frac{1}{9} \log t+C \\
& =\frac{7}{9}(3 x-2)-\frac{1}{9} \log |3 x-2|+C \\
& =\frac{7}{3} x-\frac{1}{9} \log |3 x-2|+C
\end{aligned}$
Here, $A=\frac{7}{3}, B=-\frac{1}{9}$
$\therefore \quad 3 B-A=3\left(-\frac{1}{9}\right)-\frac{7}{3}=-\frac{8}{3}$