Answer:
Let \[l=\int{{{x}^{2x}}(1+\log x)\,\,dx}\] Put \[u={{x}^{2x}}\Rightarrow \frac{du}{dx}=2{{x}^{2x}}(\log \,\,(x)+1)\] \[\Rightarrow \] \[\frac{du}{2}={{x}^{2x}}(1+\log x)\,\,dx\] \[\therefore \] \[l=\int{\frac{du}{2}=\frac{1}{2}}\int{du}\] \[=\frac{1}{2}(u)+C=\frac{u}{2}+C\] \[\Rightarrow \] \[l=\frac{{{x}^{2x}}}{2}+C\]
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