question_answer

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

Answer:

Let \[l=\int{(x+1){{e}^{x}}\log \,(x{{e}^{x}})dx}\] Put       \[x{{e}^{x}}=t\] \[\Rightarrow \]   \[(1{{e}^{x}}+x{{e}^{x}})dx=dt\] \[\Rightarrow \]   \[(1+x){{e}^{x}}dx=dt\] \[\therefore \]      \[l=\int{1\cdot \log \,\,t\,\,dt}\] \[=\log \,\,t\cdot t-\int{t\cdot \frac{1}{t}dt}\] [using integration by parts] \[=t\cdot \log t-\int{1\,dt}\] \[=t\cdot \log \,\,t-t+C\] \[=x{{e}^{x}}\log \,(x{{e}^{x}})-x{{e}^{x}}+C\] \[=x{{e}^{x}}[\log \,(x{{e}^{x}})-1]+C\]