question_answer

Evaluate \[\int{{{x}^{2x}}(1+\log \,x)dx.}\]

Answer:

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