A) \[\frac{{{x}^{n+1}}}{n+1}\left\{ \log x+\frac{1}{n+1} \right\}+c\]
B) \[\frac{{{x}^{n+1}}}{n+1}\left\{ \log x+\frac{2}{n+1} \right\}+c\]
C) \[\frac{{{x}^{n+1}}}{n+1}\left\{ 2\log x-\frac{1}{n+1} \right\}+c\]
D) \[\frac{{{x}^{n+1}}}{n+1}\left\{ \log x-\frac{1}{n+1} \right\}+c\]
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{{{x}^{n}}\log x\,dx}=\log x\,.\,\frac{{{x}^{n+1}}}{n+1}-\int_{{}}^{{}}{\frac{{{x}^{n+1}}}{n+1}\,.\,\frac{1}{x}\,dx}\] \[=\frac{{{x}^{n+1}}}{n+1}\log x-\frac{{{x}^{n+1}}}{{{(n+1)}^{2}}}+c=\frac{{{x}^{n+1}}}{n+1}\left[ \log x-\frac{1}{n+1} \right]+c.\]
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