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JEE Mains · Maths · STD 12 - 7.2 definite integral

The value of the integral \(\int\limits_{-1}^{1} \left(\dfrac{x^3 + |x| + 1}{x^2 + 2|x| + 1}\right) dx\) is equal to :

  1. A \(3\log_e 2\)
  2. B \(2\log_e 2\)
  3. C \(5\log_e 3\)
  4. D \(3\log_e 3\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2\log_e 2\)

Step-by-step Solution

Detailed explanation

Let \(I = \int_{-1}^{1} \left(\dfrac{x^3 + |x| + 1}{x^2 + 2|x| + 1}\right) dx\) We can split the integral into two parts: \(I = \int_{-1}^{1} \dfrac{x^3}{x^2 + 2|x| + 1} dx + \int_{-1}^{1} \dfrac{|x| + 1}{x^2 + 2|x| + 1} dx\) The first integrand…
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