JEE Main 2019 (12 January Evening Shift)

Question

The integral $\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$ loge x dx is equal to :

Accepted Answer

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

Answer

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

Answer

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

Answer

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

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