JEE Main 2022 · 27 June · morning · Single correct · +4 / −1

The value of the integral
∫₋₂² (|x³ + x|)/(e^(x|x|) + 1)dx is equal to :

  1. (a) 5e²
  2. (b) 3e⁻²
  3. (c) 4
  4. (d) 6
Answer and solution

Answer (d) 6

Solution

Maths sir · 4 steps

Replace x by minus x and add the two copies; the awkward exponential denominators then sum to exactly 1.

  1. Use the property. Call the integral I. Since ∫₋ₐᵃ f(x) dx = ∫₋ₐᵃ f(-x) dx, replace x by -x. The numerator |x³ + x| stays the same, and x|x| becomes -x|x|, so I = ∫₋₂² (|x³+x|)/(e^(-x|x|)+1) dx.
  2. Add the two forms. Write t = x|x|. Then 1/(eᵗ+1) + 1/(e⁻ᵗ+1) = 1/(eᵗ+1) + eᵗ/(1+eᵗ) = 1. So adding the two expressions for I removes the exponential completely.

    2I = ∫₋₂² |x³+x| dx = 2∫₀² (x³+x)dx

  3. Evaluate. The integrand |x³+x| is even, and for x ≥ 0 we have x³ + x ≥ 0. So ∫₀² (x³+x) dx = [x⁴/4 + x²/2]₀² = 4 + 2 = 6, and 2I = 12.
  4. So. I = 6, which is option (d).

The idea to keep. When a denominator looks like eᵍ⁽ˣ⁾ + 1 with g odd and the limits are symmetric, replace x by -x and add. The two fractions sum to 1, leaving only the numerator to integrate.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Definite Integration

Asked about 230 times in JEE Main since 2020. This question is from 2022.