A) \[{{x}^{x}}\]
B) \[{{x}^{2x}}\]
C) \[{{x}^{x}}\log x\]
D) \[\frac{1}{2}{{(1+\log x)}^{2}}\]
Correct Answer: A
Solution :
\[I=\int{{{x}^{x}}(1+\log x)\,dx}\]. Put \[{{x}^{x}}=t\], then \[{{x}^{x}}(1+\log x)dx=dt\] \[\therefore I=\int{dt}\] \[\Rightarrow I=t+C\] \[\Rightarrow I={{x}^{x}}+C\].
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