I=xn+1n+1log⁡x−1n+1∫xn+1⋅1xdx=xn+1n+1log⁡x−1(n+1)2xn+1 etc.

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$\int x^{n} \log x d x =$

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$\frac{x^{n + 1}}{n} [n \log x - 1]$

$\frac{x^{n + 1}}{n + 1} [(n + 1) \log x - 1]$

$\frac{x^{n + 1}}{(n + 1)^{2}} [(n + 1) \log x - 1]$

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answer is C.

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Detailed Solution

$\begin{aligned} I = \frac{x^{n + 1}}{n + 1} \log x - \frac{1}{n + 1} \int x^{n + 1} \cdot \frac{1}{x} d x \\ = \frac{x^{n + 1}}{n + 1} \log x - \frac{1}{(n + 1)^{2}} x^{n + 1} etc. \end{aligned}$

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