The integral

Question:

The integral $\int_{1}^{2} e^{x} \cdot x^{x}\left(2+\log _{e} x\right) d x$ equal :

 

  1. $e(4 e+1)$

  2. $e(2 e-1)$

  3. $4 e^{2}-1$

  4. $e(4 e-1)$


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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