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NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice

∫ ln ⁡ x + 1 - ln ⁡ x x x + 1 d x is equal to (where C is an arbitarary constant)

Options

  1. A- 1 2 ln ⁡ x + 1 x 2 + C
  2. BC - ln ⁡ x + 1 2 - ln ⁡ x 2
  3. C- ln ⁡ ln ⁡ x + 1 x + C
  4. D- ln ⁡ x + 1 x + C

Correct answer

A. - 1 2 ln ⁡ x + 1 x 2 + C

Step-by-step solution

Put ln ⁡ x + 1 - ln ⁡ x = t ⇒ 1 x + 1 - 1 x = d t d x ⇒ x - x + 1 x x + 1 = d t d x ⇒ - d x x x + 1 = d t ⇒ d x x x + 1 = - d t so question becomes - ∫ ‍ t d t = - t 2 2 + C

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