The value if the integral - 1 2 e x log e x - x-1 x dx is

The correct option is A e ^ 2 ( 1+log _ e 2 ) e Let #160; I=∫ _ 1 ^ 2 e ^ x ( log _ e x + x+1 x #160;) #160; dx ⇒ I=∫ _ 1 ^ 2 e ^ x ...

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Question

The value if the integral
$\int_{1}^{2} e^{x} ({log}_{e} x + \frac{x + 1}{x}) d x$ is

e^{2} (1 + {log}_{e} 2)

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e^{2} - e

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e^{2} (1 + {log}_{e} 2) - e

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e^{2} - e (1 + {log}_{e} 2)

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