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

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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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