The integral _1^2 0.05eme^x x^x(2+ log_e ⁡x) dx equals:

Question

Accepted Answer

Correct answer: B. (b) I= _1^2 0.05eme^xx^x(2+ log_e ⁡x) dxI= _1^2 0.05eme^xx^x(1+(1+ log_e ⁡x)) dx = _1^2 0.05eme^x(x^x+x^x(1+ log_e ⁡x)) dx = [e^xx^x ]_1^2 ,\ e^x (f (x )+f^ ' (x ) ) dx=e^xf (x )+c\ =e^2 4-e 1 =4e^2-e =e (4e-1 )

$The integral \int_{1}^{2} e^{x} \cdot x^{x} (2 + log_{e} x) dx equals:$

Solution

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