The integral


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

  1. (1) $e(4 e+1)$

  2. (2) $4 e^{2}-1$

  3. (3) $e(4 e-1)$

  4. (4) $e(2 e-1)$

Correct Option: 1


$I=\int_{1}^{2} e^{x} x^{x}\left(2+\log _{e} x\right) d x$

$I=\int_{1}^{2} e^{x} x^{x}\left[1+\left(1+\log _{e} x\right)\right] d x$

$=\int_{1}^{2} e^{x}\left[x^{x}+x^{x}\left(1+\log _{e} x\right)\right] d x$

$\because \int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c$

$\therefore I=\left[e^{x} x^{x}\right]_{1}^{2}=e^{2} \times 4-e \times 1=4 e^{2}-e=e(4 e-1)$

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