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 :

Question

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

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

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

${1 \over 2} - e - {1 \over {{e^2}}}$

${3 \over 2} - {1 \over e} - {1 \over {2{x^2}}}$

Solution

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