If {I_n} = int {{(log x)}^n} , dx,then {I_n} | vedclass

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(A) Given ${I_n} = \int {{(\log x)}^n} \, dx$. Using integration by parts,let $u = {(\log x)^n}$ and $dv = dx$. Then $du = n{(\log x)}^{n - 1} \cdot \

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$x{(\log x)^n}$

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(A) Given ${I_n} = \int {{(\log x)}^n} \, dx$. Using integration by parts,let $u = {(\log x)^n}$ and $dv = dx$. Then $du = n{(\log x)}^{n - 1} \cdot \frac{1}{x} \, dx$ and $v = x$. Applying the formula $\int u \, dv = uv - \int v \, du$: ${I_n} = x{(\log x)^n} - \int x \cdot n{(\log x)}^{n - 1} \cdot \frac{1}{x} \, dx$ ${I_n} = x{(\log x)^n} - n \int {{(\log x)}^{n - 1}} \, dx$ Since ${I_{n - 1}} = \int {{(\log x)}^{n - 1}} \, dx$,we have: ${I_n} = x{(\log x)^n} - n{I_{n - 1}}$ Rearranging the terms,we get: ${I_n} + n{I_{n - 1}} = x{(\log x)^n}$.

If ${I_n} = \int {{(\log x)}^n} \, dx$,then ${I_n} + n{I_{n - 1}} = $

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