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$I f I_{n} = \int {(\ln x)}^{n} d x t h e n I_{n} + n I_{n - 1} =$

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(ln x)nx

x(ln x)n-1

x(ln x)n

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Correct option is C

In=∫ln xn.1 dxIntegrating by parts =xln xn−∫x n ln xn−1xdx =xln xn−nIn−1∴In+nIn−1=xln xn

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If In=∫(ln x)ndx then In+nIn-1 =

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$I f I_{n} = \int {(\ln x)}^{n} d x t h e n I_{n} + n I_{n - 1} =$