The integral int_{1}^{e} left( left( frac{x}{ | vedclass

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(D) Let $I = \int_{1}^{e} \left( \left( \frac{x}{e} \right)^{2x} - \left( \frac{e}{x} \right)^{x} \right) \log_{e} x \, dx$. Split the integral: $I =

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$\frac{3}{2} - e - \frac{1}{2e^2}$

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(D) Let $I = \int_{1}^{e} \left( \left( \frac{x}{e} \right)^{2x} - \left( \frac{e}{x} \right)^{x} \right) \log_{e} x \, dx$. Split the integral: $I = \int_{1}^{e} \left( \frac{x}{e} \right)^{2x} \log_{e} x \, dx - \int_{1}^{e} \left( \frac{e}{x} \right)^{x} \log_{e} x \, dx$. For the first integral,let $u = \left( \frac{x}{e} \right)^{2x}$. Then $\log_{e} u = 2x \log_{e} \left( \frac{x}{e} \right) = 2x (\log_{e} x - 1)$. Differentiating both sides: $\frac{1}{u} du = 2(\log_{e} x - 1) dx + 2x \cdot \frac{1}{x} dx = 2 \log_{e} x \, dx$. So,$\log_{e} x \, dx = \frac{1}{2} \frac{du}{u}$. When $x=1, u = (1/e)^2 = e^{-2}$. When $x=e, u = (e/e)^{2e} = 1$. Thus,$\int_{1}^{e} \left( \frac{x}{e} \right)^{2x} \log_{e} x \, dx = \int_{e^{-2}}^{1} u \cdot \frac{1}{2} \frac{du}{u} = \frac{1}{2} [u]_{e^{-2}}^{1} = \frac{1}{2} (1 - e^{-2})$. For the second integral,let $v = \left( \frac{e}{x} \right)^{x}$. Then $\log_{e} v = x \log_{e} \left( \frac{e}{x} \right) = x (1 - \log_{e} x)$. Differentiating: $\frac{1}{v} dv = (1 - \log_{e} x) dx + x (-1/x) dx = -\log_{e} x \, dx$. So,$\log_{e} x \, dx = -\frac{dv}{v}$. When $x=1, v = (e/1)^1 = e$. When $x=e, v = (e/e)^e = 1$. Thus,$\int_{1}^{e} \left( \frac{e}{x} \right)^{x} \log_{e} x \, dx = \int_{e}^{1} v \cdot (-dv/v) = -\int_{e}^{1} dv = [v]_{1}^{e} = e - 1$. Combining these: $I = \frac{1}{2} (1 - e^{-2}) - (e - 1) = \frac{1}{2} - \frac{1}{2e^2} - e + 1 = \frac{3}{2} - e - \frac{1}{2e^2}$.

The integral $\int_{1}^{e} \left( \left( \frac{x}{e} \right)^{2x} - \left( \frac{e}{x} \right)^{x} \right) \log_{e} x \, dx$ is equal to

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