AP EAMCET202119 Aug 2021Morning ShiftMathematicsDefinite IntegrationActual
∫ − 1 / 2 1 / 2   [ x ] + log ⁡ 1 + x 1 − x d x =
Options
- A2 log ⁡ ( 1 / 2 )
- B0
- C− 1 2
- D1
Correct answer
C. − 1 2
Step-by-step solution
Given that ∫ − 1 / 2 1 / 2   [ x ] + log ⁡ 1 + x 1 − x d x Let f x   =   log 1 + x 1 - x   ⇒ f - x   =   log 1 - x 1 + x f - x   =   log 1 + x 1 - x - 1 ⇒ f - x   =   - log 1 + x 1 - x ⇒ f - x   = - f x   So f x is an odd function ⇒ ∫ - 1 2 1 2 f x d x   =   0   ;   f - x   = - f x   ∫ - 1 2 1 2 x d x   =   ∫ - 1 2 0 x d x   +   ∫ 0 1