If $f\left(x\right)=\frac{3-8x}{3x-1}$ and $\int f\left(y\right)dy=Ay+B \log \left|3y-1\right|+C$, then $\frac{A-3B}{2}=$

If $f\left(x\right)=\frac{3-8x}{3x-1}$ and $\int f\left(y\right)dy=Ay+B \log \left|3y-1\right|+C$ 

Now, $\int \frac{3-8y}{3y-1}dy=Ay+B\log \left|3y-1\right|+C.............\left(i\right)$

$=\int \frac{3}{3y-1}dy-\int \frac{8y}{3y-1}dy$

$=\int \frac{3}{3y-1}dy-\frac{8}{3}\int \frac{3y}{3y-1}dy$

$=\int \frac{3}{3y-1}dy-\frac{8}{3}\int \frac{3y-1+1}{3y-1}dy$

$=\int \frac{3}{3y-1}dy-\frac{8}{3}\int \frac{3y-1}{3y-1}dy-\frac{8}{3}\int \frac{1}{3y-1}dy$

$=\frac{3}{3}\log \left|3y-1\right|-\frac{8}{3}y-\frac{8}{3\times 3}\log \left|3y-1\right|+C$

$\log \left|3y-1\right|-\frac{8}{3}y-\frac{8}{9}\log \left|3y-1\right|+C$

From equation $\left(i\right)$

$\frac{1}{9}\log \left|3y-1\right|-\frac{8}{3}y+C=Ay+B\log \left|3y-1\right|+C$

Comparing on both sides

$A=-\frac{8}{3}, B=\frac{1}{9}$

Now, putting values of $A\&B$

$\frac{A-3B}{2}=\frac{-{\displaystyle \frac{8}{3}}-3\times {\displaystyle \frac{1}{9}}}{2}$

$\frac{A-3B}{2}=\frac{-3}{2}$