For I_n = int_{1}^{e} (ln x)^n dx,where n in | vedclass

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(C) Given $I_n = \int_{1}^{e} (\ln x)^n dx$. Using integration by parts,let $u = (\ln x)^n$ and $dv = dx$. Then $du = n(\ln x)^{n-1} \cdot \frac{1}{x}

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$I_{n + 1} + (n + 1) I_n = e$

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(C) Given $I_n = \int_{1}^{e} (\ln x)^n dx$. Using integration by parts,let $u = (\ln x)^n$ and $dv = dx$. Then $du = n(\ln x)^{n-1} \cdot \frac{1}{x} dx$ and $v = x$. $I_n = [x(\ln x)^n]_{1}^{e} - \int_{1}^{e} x \cdot n(\ln x)^{n-1} \cdot \frac{1}{x} dx$. $I_n = (e(\ln e)^n - 1(\ln 1)^n) - n \int_{1}^{e} (\ln x)^{n-1} dx$. Since $\ln e = 1$ and $\ln 1 = 0$,we have $I_n = e - n I_{n-1}$. Replacing $n$ with $n+1$,we get $I_{n+1} = e - (n+1) I_n$. Rearranging the terms,we get $I_{n+1} + (n+1) I_n = e$.

For $I_n = \int_{1}^{e} (\ln x)^n dx$,where $n \in N$,which of the following relations holds true?

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