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MHT CET20192 May 2019Morning ShiftMathematicsDefinite IntegrationActual

∫ 0 4 1 1 + x d x = ________

Options

  1. Alog ⁡ e 4 6
  2. Blog ⁡ e 4 3
  3. Clog ⁡ e 4 9
  4. Dlog ⁡ e 3 4

Correct answer

C. log ⁡ e 4 9

Step-by-step solution

Let I = ∫ 0 4 d x 1 + x put x = t 2 d x = 2 t d t Then, I = ∫ 0 2 2 t d t 1 + t = 2 ∫ 0 2 1 + t - 1 1 + t d t I = 2 ∫ 0 2 1 - 1 1 + t d t I = 2 2 - l n 3 = l o g e 4 9

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