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JEE Main202120 Jul 2021Morning ShiftMathematicsDefinite IntegrationActual

The value of the integral ∫ - 1 1 log e 1 - x + 1 + x d x is equal to:

Options

  1. A1 2 log e 2 + π 4 - 3 2
  2. B2 log e 2 + π 4 - 1
  3. Clog e 2 + π 2 - 1
  4. D2 log e 2 + π 2 - 1 2

Correct answer

C. log e 2 + π 2 - 1

Step-by-step solution

Let I = ∫ - 1 1 log e 1 - x + 1 + x d x ⇒ I = 2 ∫ 0 1 log e 1 - x + 1 + x ⏟ I · 1 ⏟ I I d x ⇒ I = 2 x log e 1 - x + 1 + x 0 1 - ∫ 0 1 x · 1 1 - x + 1 + x · 1 2 1 + x - 1 2 1 - x d x ⇒ I = 2 log e 2 - 2 2 ∫ 0 1 x 1 - x - 1 + x 1 - x + 1 + x 1 - x 2 d x ⇒ I = 2 log e 2 - ∫ 0 1 x 1 - x - 1 + x 2 - 2 x 1 - x 2 d x ⇒ I = log e 2 + 1 2 ∫ 0 1 2 - 2 1 - x 2 1 - x 2   d x ⇒ I = log e 2 + ∫ 0 1 1 - 1 - x 2 1 - x 2 d

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