Mathematics · 2019
JEE Main · 12 January 2019, Shift 2 · Q75
The integral ∫_1^e{(x/e)^2 x-(e/x)^x} log_e x d x is equal to:
The integral $\displaystyle \int_1^e\left\{\left(\frac{x}{e}\right)^{2 x}-\left(\frac{e}{x}\right)^x\right\} \log _e x d x$
is equal to :
Official answer
(2)
$\displaystyle \frac{3}{2}-e-\frac{1}{2 e^2}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.