imo class 12 definite integration

Evaluate ∫₁ᵉ (1 + log x) dx.

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Evaluate ∫₁ᵉ (1 + log x) dx.

  • A. e
  • B. e + 1
  • C. e − 1
  • D. 1

Answer: A) e

Explanation: ∫₁ᵉ (1 + log x) dx = [x + x log x − x]₁ᵉ? Actually, ∫ log x dx = x log x − x. So ∫(1+log x)dx = x + x log x − x = x log x. From 1 to e: e log e − 1 log 1 = e − 0 = e.

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