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JEE Main 2019 · 12 Jan · Shift II

JEE Main 2019 Mathematics question on Definite Integrals and Areas

The integral ∫1e[(xe)2x−(ex)x]log⁡ex dx\int_1^{e}\left[\left(\frac{x}{e}\right)^{2x} - \left(\frac{e}{x}\right)^x\right]\log_e x\, dx is equal to
  1. A.
    32−e−12e2\frac{3}{2} - e - \frac{1}{2e^2}
  2. B.
    −12+1e−12e2-\frac{1}{2} + \frac{1}{e} - \frac{1}{2e^2}
  3. C.
    12−e−1e2\frac{1}{2} - e - \frac{1}{e^2}
  4. D.
    32−1e−12e2\frac{3}{2} - \frac{1}{e} - \frac{1}{2e^2}
Show answer

Correct option: A

[JEE Main 2019 · 12 Jan · Shift II]

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