JEE Main202126 Aug 2021Morning ShiftMathematicsDefinite IntegrationActual
The value of ∫ - 1 2 1 2 x + 1 x - 1 2 + x - 1 x + 1 2 - 2 1 2 d x is:
Options
- Alog e 4
- B2 log e 16
- Clog e 16
- D4 log e ( 3 + 2 2 )
Correct answer
C. log e 16
Step-by-step solution
We have, ∫ - 1 2 1 2 x + 1 x - 1 2 + x - 1 x + 1 2 - 2 1 2   d x = ∫ - 1 2 1 2 x - 1 x + 1 - x + 1 x - 1 2 d x = ∫ - 1 2 1 2 - 4 x x 2 - 1 2 d x = ∫ - 1 2 1 2 4 x 1 - x 2 2 d x = ∫ - 1 2 1 2 16 x 2 1 - x 2 2 d x = ∫ - 1 2 1 2 4 x 1 - x 2 d x = 2 ∫ 0 1 2 4 x 1 - x 2 d x = 4 ∫ 0 1 2 2 x 1 - x 2 d x = - 4 ∫ 0 1 2 - 2 x 1 - x 2 d x = - 4 log e 1 - x 2 0 1 2 = - 4 log e 1 - 1 2 = 4 log e 2 = log e 16