JEE Main2015MathematicsDefinite IntegrationActual
For x > 0 , let f x = ∫ 1 x log ⁡ t 1 + t d t . Then f x + f 1 x is equal to
Options
- A1 2 log x 2
- Blog x
- C1 4 log x 2
- D1 4 log x 2
Correct answer
A. 1 2 log x 2
Step-by-step solution
f x =   ∫ 1 x log ⁡ t 1 + t   d t ⇒ f x = ∫ 1 x log ⁡ z ( 1 + z ) d z And f 1 x =   ∫ 1 1 x log ⁡ t 1 + t   d t Put t = 1 z d t =   - 1 z 2   d z f x =   ∫ 1 x log ⁡ z z 2 1 +   1 z   ⋅ d z f x =   ∫ 1 x log ⁡ z z ( 1 + z )     d z f x + f 1 x =   ∫ 1 x log ⁡ z   1 1 + z + 1 z ( 1 + z )   d z =   ∫ 1 x 1 z log ⁡ z   d z Put log ⁡ z