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If $J_{m} = \int_{1}^{e} \log^{m} x d x$ then $J_{8} + 8 J_{7}$ is equal to

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detailed solution

Correct option is D

Jm=∫1e logm⁡xdx=xlogm⁡x1e−∫1e mxlogm−1⁡xxdx=e−mJm−1So J8+8J7=e

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If Jm=∫1e logm⁡xdxthen J8+8J7 is equal to

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If $J_{m} = \int_{1}^{e} \log^{m} x d x$ then $J_{8} + 8 J_{7}$ is equal to