If n in N and I_{n}=int(log x)^{n} dx,then I_ | vedclass

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(B) 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} d

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

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(B) 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 $\int u dv = uv - \int v du$: $I_{n} = x(\log x)^{n} - \int x \cdot n(\log x)^{n-1} \cdot \frac{1}{x} dx$ $I_{n} = x(\log x)^{n} - n \int (\log x)^{n-1} dx$ $I_{n} = x(\log x)^{n} - n I_{n-1}$ Rearranging the terms,we get: $I_{n} + n I_{n-1} = x(\log x)^{n} + C$.

If $n \in N$ and $I_{n}=\int(\log x)^{n} dx$,then $I_{n}+n I_{n-1}$ is equal to

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