Official solution
\[\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx\]
= \[\int\limits_1^2 {{e^x}{x^x}\left[ {1 + \left( {1 + {{\log }_e}x} \right)} \right]} dx\]
= \[\int\limits_1^2 {{e^x}\left[ {{x^x} + {x^x}\left( {1 + {{\log }_e}x} \right)} \right]} dx\]
= \[\left[ {{e^x}{x^x}} \right]_1^2\]
= e2 \[ \times \] 4 - e \[ \times \] 1
= 4e2 - e
= e(4e - 1)
Note : \[\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} \] = exf(x) + c