NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice

∫ ln x - 1 x + 1 x 2 - 1 dx is equal to

Options

A1 2 ln x - 1 x + 1 2 + C
B1 2 ln x + 1 x - 1 2 + C
C1 4 ln x - 1 x + 1 2 + C
D1 4  ln  x + 1 x - 1 +   C

Correct answer

C. 1 4 ln x - 1 x + 1 2 + C

Step-by-step solution

I = ∫ ln x - 1 x + 1 x 2 - 1 dx , Let t = ln x - 1 x + 1 ⇒ dt dx = x + 1 x - 1 x + 1 - x - 1 x + 1 2 = 2 x 2 - 1 ⇒ dx x 2 - 1 = dt 2 ⇒ I = 1 2 ∫ tdt = 1 4 t 2 + C = 1 4 ln x - 1 x + 1 2 + C

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