If int frac{(2x+1)^6}{(3x+2)^8} dx = P left( | vedclass

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

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$\frac{1}{7^2}$

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(A) Given,$\int \frac{(2x+1)^6}{(3x+2)^8} dx = P \left( \frac{2x+1}{3x+2} \right)^Q + R$ Let $t = \frac{2x+1}{3x+2}$. Then,$\frac{dt}{dx} = \frac{(3x+2)(2) - (2x+1)(3)}{(3x+2)^2} = \frac{6x+4-6x-3}{(3x+2)^2} = \frac{1}{(3x+2)^2}$. Thus,$dt = \frac{dx}{(3x+2)^2}$. The integral becomes $\int \left( \frac{2x+1}{3x+2} \right)^6 \cdot \frac{dx}{(3x+2)^2} = \int t^6 dt$. Integrating,we get $\int t^6 dt = \frac{t^7}{7} + C = \frac{1}{7} \left( \frac{2x+1}{3x+2} \right)^7 + C$. Comparing this with $P \left( \frac{2x+1}{3x+2} \right)^Q + R$,we get $P = \frac{1}{7}$,$Q = 7$,and $R = C$. Therefore,$\frac{P}{Q} = \frac{1/7}{7} = \frac{1}{49} = \frac{1}{7^2}$.

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

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