Question:
The integral $\int_{1}^{2} e^{x} \cdot x^{x}\left(2+\log _{e} x\right) d x$ equals:
Correct Option: 1
Solution:
$I=\int_{1}^{2} e^{x} x^{x}\left(2+\log _{e} x\right) d x$
$I=\int_{1}^{2} e^{x} x^{x}\left[1+\left(1+\log _{e} x\right)\right] d x$
$=\int_{1}^{2} e^{x}\left[x^{x}+x^{x}\left(1+\log _{e} x\right)\right] d x$
$\because \int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c$
$\therefore I=\left[e^{x} x^{x}\right]_{1}^{2}=e^{2} \times 4-e \times 1=4 e^{2}-e=e(4 e-1)$