JEE Math Practice 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 :
- A.$$ - {1 \over 2} + {1 \over e} - {1 \over {2{e^2}}}$$
- B.$${3 \over 2} - e - {1 \over {2{e^2}}}$$
- C.$${1 \over 2} - e - {1 \over {{e^2}}}$$
- D.$${3 \over 2} - {1 \over e} - {1 \over {2{x^2}}}$$
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