∫_1^e e^x/x(1+x log x)dx=∫_1^e 1/xe^x dx+∫_1^e e^x log_ex dx =[e^xlog x]_1^e ∫_1^e e^x log x dx+∫_1^e e^x log x dx =[e^e log e e^1 log_e 1]=e^e ...

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First Fundamental Theorem of Calculus

∫1e ex/x1+x l...

Question

$\int_{1}^{e} \frac{e^{x}}{x} (1 + x l o g x) d x =$

e^{e}

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e^{e} - e

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e^{e} + e

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N o n e o f t h e s e

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Solution

The correct option is A $e^{e}$
$\int_{1}^{e} \frac{e^{x}}{x} (1 + x l o g x) d x = \int_{1}^{e} \frac{1}{x} e^{x} d x + \int_{1}^{e} e^{x} l o g_{e} x d x$
$= [e^{x} l o g x]_{1}^{e} - \int_{1}^{e} e^{x} l o g x d x + \int_{1}^{e} e^{x} l o g x d x$
$= [e^{e} l o g e - e^{1} l o g_{e} 1] = e^{e}$

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