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

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(A) આપેલ છે,$\int \frac{(2x+1)^6}{(3x+2)^8} dx = P \left( \frac{2x+1}{3x+2} \right)^Q + R$ ધારો કે $t = \frac{2x+1}{3x+2}$. તેથી,$\frac{dt}{dx} = \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) આપેલ છે,$\int \frac{(2x+1)^6}{(3x+2)^8} dx = P \left( \frac{2x+1}{3x+2} \right)^Q + R$ ધારો કે $t = \frac{2x+1}{3x+2}$. તેથી,$\frac{dt}{dx} = \frac{(3x+2)(2) - (2x+1)(3)}{(3x+2)^2} = \frac{6x+4-6x-3}{(3x+2)^2} = \frac{1}{(3x+2)^2}$. આમ,$dt = \frac{dx}{(3x+2)^2}$. સંકલન આ મુજબ થશે: $\int \left( \frac{2x+1}{3x+2} \right)^6 \cdot \frac{dx}{(3x+2)^2} = \int t^6 dt$. સંકલન કરતા,$\int t^6 dt = \frac{t^7}{7} + C = \frac{1}{7} \left( \frac{2x+1}{3x+2} \right)^7 + C$. $P \left( \frac{2x+1}{3x+2} \right)^Q + R$ સાથે સરખાવતા,આપણને $P = \frac{1}{7}$,$Q = 7$ અને $R = C$ મળે છે. તેથી,$\frac{P}{Q} = \frac{1/7}{7} = \frac{1}{49} = \frac{1}{7^2}$.

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

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