A) \[{{x}^{x}}+c\]
B) \[x\cdot \log x+c\]
C) \[{{(\log x)}^{x}}+c\]
D) \[{{x}^{\log x}}+c\]
Correct Answer: A
Solution :
Let\[I=\int{{{x}^{x}}}(\log ex)dx\] \[=\int{{{x}^{x}}}(1+\log x)dx\] let\[t={{x}^{x}}={{e}^{x\log x}}\] \[\Rightarrow \]\[\frac{dt}{dx}={{x}^{x}}\left\{ {{x}^{x}}\cdot \frac{1}{x}+\log x \right\}\] \[\Rightarrow \,\,\,dt={{x}^{x}}(1+\log x)dx\] \[\therefore \,\,\,\,\,l=\int_{{}}^{{}}{t+c={{x}^{x}}+c}\]
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