If ∫ ( 2 x + 1 ) 6 ( 3 x + 2 ) 8 d x = P 2 x + 1 3 x + 2 Q + R , then P Q =

Given, $\int \frac{(2x+1{)}^6}{(3x+2{)}^8}dx=P{\left(\frac{2x+1}{3x+2}\right)}^Q+R$

Put $t=\frac{2x+1}{3x+2}$

$\Rightarrow \frac{\mathit{d}t}{\mathit{d}x}=\frac{\left(3x+2\right)\left(2\right)-\left(2x+1\right)\left(3\right)}{{\left(3x+2\right)}^2}$

$\Rightarrow \mathit{d}t=\frac{\mathit{d}x}{{\left(3x+2\right)}^2}$

$\therefore \int \frac{(2x+1{)}^6}{(3x+2{)}^8}dx=\int t^{6}dt$

$\int t^{6}dt=\frac{t^7}{7}+C=\frac{1}{7}{\left(\frac{2x+1}{3x+2}\right)}^7+C$

$\Rightarrow P=\frac{1}{7}, Q=7 \& R=C$

$\therefore \frac{P}{Q}=\frac{{\displaystyle \frac{1}{7}}}{7}=\frac{1}{7^2}$