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JEE Main20206 Sep 2020Evening ShiftMathematicsDefinite IntegrationActual

The integral ∫ 1 2 e x . x x 2 + log e x dx equals :

Options

  1. Ae 4 e + 1
  2. B4 e 2 - 1
  3. Ce 4 e - 1
  4. De 2 e - 1

Correct answer

C. e 4 e - 1

Step-by-step solution

Let y = ex x l ny = x 1 + l nx 1 y dy dx = 2 + l nx ⇒ dy = ex x 2 + l nx   dx ∫ 1 2 e x . x x 2 + log e x dx = y 1 2 = ex x 1 2 = 4 e 2 − e

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