-1e e x-x1-x log x d x-A- e e - eB- e e - eC- None of theseD- e e

The correct option is D e^e∫_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 l ...

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Byju's Answer

Standard XII

Mathematics

Geometrical Interpretation

∫1e e x/x1+x ...

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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Similar questions

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(b) 1/e

(c)

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(d) 2/e

(e)

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Q. Evaluate

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Q. Solve:

x^{e} + e^{x} + e^{e}

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∫e1exx(1+x log x)dx=

The correct option is A ee∫e1exx(1+x log x)dx=∫e11xex dx+∫e1 ex logex dx =[exlog x]e1−∫e1ex log x dx+∫e1exlog x dx =[ee log e−e1loge1]=ee

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

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