Answer:
We have, \[l=\int{{{x}^{x}}(1+\log \,\,x)\,dx}\] Put \[{{x}^{x}}=t\] \[\Rightarrow \] \[{{x}^{x}}(1+\log \,\,x)\,dx=dt\] \[\therefore \] \[l=\int{dt=t+C={{x}^{x}}+C}\]
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