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

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(A) दिया गया है,$\int \frac{(2x+1)^6}{(3x+2)^8} dx = P \left( \frac{2x+1}{3x+2} \right)^Q + R$ माना $t = \frac{2x+1}{3x+2}$। तब,$\frac{dt}{dx} = \frac

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

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(A) दिया गया है,$\int \frac{(2x+1)^6}{(3x+2)^8} dx = P \left( \frac{2x+1}{3x+2} \right)^Q + R$ माना $t = \frac{2x+1}{3x+2}$। तब,$\frac{dt}{dx} = \frac{(3x+2)(2) - (2x+1)(3)}{(3x+2)^2} = \frac{6x+4-6x-3}{(3x+2)^2} = \frac{1}{(3x+2)^2}$। अतः,$dt = \frac{dx}{(3x+2)^2}$। समाकलन इस प्रकार होगा: $\int \left( \frac{2x+1}{3x+2} \right)^6 \cdot \frac{dx}{(3x+2)^2} = \int t^6 dt$। समाकलन करने पर,$\int t^6 dt = \frac{t^7}{7} + C = \frac{1}{7} \left( \frac{2x+1}{3x+2} \right)^7 + C$। $P \left( \frac{2x+1}{3x+2} \right)^Q + R$ के साथ तुलना करने पर,हमें $P = \frac{1}{7}$,$Q = 7$ और $R = C$ प्राप्त होता है। इसलिए,$\frac{P}{Q} = \frac{1/7}{7} = \frac{1}{49} = \frac{1}{7^2}$।

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

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