For n in N,let P_n = int_1^e (ln x)^n dx. The | vedclass

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(B) Given $P_n = \int_1^e (\ln x)^n dx$. Using integration by parts,let $u = (\ln x)^n$ and $dv = dx$. Then $du = n(\ln x)^{n-1} \cdot \frac{1}{x} dx$

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$-9e$

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(B) Given $P_n = \int_1^e (\ln x)^n dx$. 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 $v = x$. $P_n = [x(\ln x)^n]_1^e - \int_1^e x \cdot n(\ln x)^{n-1} \cdot \frac{1}{x} dx$. $P_n = e(1)^n - 0 - n \int_1^e (\ln x)^{n-1} dx$. $P_n = e - n P_{n-1}$. Now,we need to find $P_{10} - 90P_8$. From the recurrence relation: $P_{10} = e - 10P_9$. $P_9 = e - 9P_8$. Substitute $P_9$ into the expression for $P_{10}$: $P_{10} = e - 10(e - 9P_8)$. $P_{10} = e - 10e + 90P_8$. $P_{10} = -9e + 90P_8$. Therefore,$P_{10} - 90P_8 = -9e$.

For $n \in N$,let $P_n = \int_1^e (\ln x)^n dx$. Then $(P_{10} - 90P_8)$ is equal to

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