The value of ∫1e((ex)2x+(xe)x)logexdx is equal to (A) e−2e21−21 (B) e−2e21+21 (C) e3−2e21−21 (D) none of these

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The value of {_{{1}}^{{e}}}{({({{x}}{{e}})}^{{{2}{x}}}+{({{e}}

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The value of $\int_{1}^{e} ((\frac{x}{e})^{2 x} + (\frac{e}{x})^{x}) lo g_{e} x d x$ is equal to (A) $e - \frac{1}{2 e ^{2}} - \frac{1}{2}$ (B) $e - \frac{1}{2 e ^{2}} + \frac{1}{2}$ (C) $e^{3} - \frac{1}{2 e ^{2}} - \frac{1}{2}$ (D) none of these

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James Discussed

The value of

\int_{1}^{e} ((\frac{x}{e})^{2 x} + (\frac{e}{x})^{x}) lo g_{e} x d x

is equal to (A)

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

(B)

e - \frac{1}{2 e ^{2}} + \frac{1}{2}

(C)

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

(D) none of these

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Question Text	The value of $\int_{1}^{e} ((\frac{x}{e})^{2 x} + (\frac{e}{x})^{x}) lo g_{e} x d x$ is equal to (A) $e - \frac{1}{2 e ^{2}} - \frac{1}{2}$ (B) $e - \frac{1}{2 e ^{2}} + \frac{1}{2}$ (C) $e^{3} - \frac{1}{2 e ^{2}} - \frac{1}{2}$ (D) none of these
Answer Type	Video solution: 1
Upvotes	150