Question:
The integral $\int_{1}^{2} e^{x} \cdot x^{x}\left(2+\log _{e} x\right) d x$ equal :
Correct Option: , 4
Solution:
$\int_{1}^{2} e^{x} \cdot x^{x}\left(2+\log _{e} x\right) d x$
$\int_{1}^{2} e^{x}\left(2 x^{x}+x^{x} \log _{e} x\right) d x$
$\int_{1}^{2} \mathrm{e}^{\mathrm{x}}(\underbrace{\mathrm{x}^{\mathrm{x}}}_{f(\mathrm{x})}+\underbrace{\mathrm{x}^{\mathrm{x}}\left(1+\log _{\mathrm{e}} \mathrm{x}\right)}_{f^{\prime}(\mathrm{x})}) d \mathrm{x}$
$\left(e^{x} \cdot x^{x}\right)_{1}^{2}=4 e^{2}-e$
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