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Let Ι 1 = ∫ 0 1 ln ⁡ x x 2 + 4 x + 1 d x and Ι 2 = ∫ 1 ∞ ln ⁡ x x 2 + 4 x + 1 d x , then

Options

  1. AΙ 1 = Ι 2
  2. BΙ 1 > Ι 2
  3. CΙ 1 + Ι 2 = 0
  4. DΙ 1 = 2 Ι 2

Correct answer

A. Ι 1 = Ι 2

Step-by-step solution

In Ι 2 substitute x = 1 t ⇒ d x = - d t t 2 So, Ι 2 = - ∫ 1 0 ln ⁡ t t 2 + 4 t + 1 ⋅ t 2 - d t t 2 = ∫ 0 1 - ln ⁡ t t 2 + 4 t + 1 d t = Ι 1 a s   ln ⁡ t = - ln ⁡ t   i n   0,1

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