∫_1^t e^x/x(1+x log x)dx=∫_1^t 1/xe^x dx + ∫_1^x e^x log_e x dx = [ e^x log x ]^4_1 ∫_1^te^x log x dx + ∫_1^t e^x log x dx =[e^t log e e^1 log_4 1]=e^t ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

∫1t ex/x1+x l...

Question

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

e^{t}

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e^{t} - e

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e^{t} + 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 $e^{t}$
$\int_{1}^{t} \frac{e^{x}}{x} (1 + x l o g x) d x = \int_{1}^{t} \frac{1}{x} e^{x} d x + \int_{1}^{x} e^{x} l o g_{e} x d x$
$= {[e^{x} l o g x]}_{1}^{x} - \int_{1}^{t} e^{x} l o g x d x + \int_{1}^{t} e^{x} l o g x d x$
$= [e^{t} l o g e - e^{1} l o g_{4} 1] = e^{t}$

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∫t1 exx(1+x log x)dx=

The correct option is A et∫t1 exx(1+x log x)dx=∫t11xexdx+∫x1ex loge x dx =[ex log x]41−∫t1ex log x dx+∫t1ex log x dx =[et log e−e1 log41]=et

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