int_1^{e} frac{e^x}{x}(1+x log x) d x= | vedclass

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(A) Let $I = \int_1^{e} \frac{e^x}{x}(1+x \log x) d x$. We can rewrite the integrand as: $I = \int_1^{e} (\frac{e^x}{x} + e^x \log x) d x$. Let $f(x)

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$e^e$

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(A) Let $I = \int_1^{e} \frac{e^x}{x}(1+x \log x) d x$. We can rewrite the integrand as: $I = \int_1^{e} (\frac{e^x}{x} + e^x \log x) d x$. Let $f(x) = e^x \log x$. Then $f'(x) = e^x \log x + e^x \cdot \frac{1}{x} = e^x (\log x + \frac{1}{x})$. This matches the integrand exactly. Therefore,$\int (f(x) + f'(x)) d x = f(x) + C$. So,$I = [e^x \log x]_1^e$. Evaluating the limits: $I = (e^e \log e) - (e^1 \log 1)$. Since $\log e = 1$ and $\log 1 = 0$,$I = (e^e \cdot 1) - (e \cdot 0) = e^e$. The correct option is $A$.

$\int_1^{e} \frac{e^x}{x}(1+x \log x) d x=$

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$\int_1^e {\frac{{{e^x}}}{x}(1 + x\log x)\,dx} = $

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