int_1^e {frac{{{e^x}}}{x}(1 + xlog x),dx} = | vedclass

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(A) We are given the integral $I = \int_1^e \frac{e^x}{x}(1 + x \log x) \, dx$. Expanding the integrand,we get: $I = \int_1^e \frac{e^x}{x} \, dx + \i

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$e^e$

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(A) We are given the integral $I = \int_1^e \frac{e^x}{x}(1 + x \log x) \, dx$. Expanding the integrand,we get: $I = \int_1^e \frac{e^x}{x} \, dx + \int_1^e e^x \log x \, dx$. Using integration by parts on the second term $\int_1^e e^x \log x \, dx$,let $u = \log x$ and $dv = e^x \, dx$. Then $du = \frac{1}{x} \, dx$ and $v = e^x$. Applying the formula $\int u \, dv = uv - \int v \, du$: $\int_1^e e^x \log x \, dx = [e^x \log x]_1^e - \int_1^e \frac{e^x}{x} \, dx$. Substituting this back into the expression for $I$: $I = \int_1^e \frac{e^x}{x} \, dx + ([e^x \log x]_1^e - \int_1^e \frac{e^x}{x} \, dx)$. The terms $\int_1^e \frac{e^x}{x} \, dx$ cancel out. $I = [e^x \log x]_1^e = (e^e \log e - e^1 \log 1)$. Since $\log e = 1$ and $\log 1 = 0$,we have: $I = e^e(1) - e(0) = e^e$.

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

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