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Mathematics · Definite Integration

JEE Main 2025 — 2 April, Evening Shift — Question 37

4∫01(13+x2+1+x2)dx−3log⁡e(3)4 \int_{0}^{1}\left(\frac{1}{\sqrt{3+x^{2}}+\sqrt{1+x^{2}}}\right) d x-3 \log _{e}(\sqrt{3}) is equal to:

  1. Option A:

    2+2−log⁡e(1+2)2+\sqrt{2}-\log _{e}(1+\sqrt{2})

  2. Option B:

    2−2−log⁡e(1+2)2-\sqrt{2}-\log _{e}(1+\sqrt{2})

    Correct
  3. Option C:

    2+2+log⁡e(1+2)2+\sqrt{2}+\log _{e}(1+\sqrt{2})

  4. Option D:

    2−2+log⁡e(1+2)2-\sqrt{2}+\log _{e}(1+\sqrt{2})

Answer: B

Step-by-step solution

Rationalize: 1/(sqrt(3+x^2)+sqrt(1+x^2)) = (sqrt(3+x^2)-sqrt(1+x^2))/2. So I = 2∫0^1 [sqrt(3+x^2)-sqrt(1+x^2)] dx. Using ∫sqrt(a^2+x^2) dx = (x/2)sqrt(a^2+x^2)+(a^2/2)ln(x+sqrt(a^2+x^2)), evaluate: ∫0^1 sqrt(3+x^2) dx = 1 + (3/2)ln(2+? wait: at x=1, sqrt(4)=2, so term = 1 + (3/2)ln(2+sqrt(4)? Actually sqrt(3+1)=2, so ln(1+2)=ln3; at 0, (3/2)ln(sqrt3). So = 1 + (3/2)ln3 - (3/2)ln(sqrt3) = 1 + (3/2)ln(3/sqrt3)=1+(3/2)ln(sqrt3)=1+(3/4)ln3. ∫0^1 sqrt(1+x^2) dx = (1/2)(sqrt2 + ln(1+sqrt2)). Then I = 2[1+(3/4)ln3 - (1/2)(sqrt2+ln(1+sqrt2))] = 2 + (3/2)ln3 - sqrt2 - ln(1+sqrt2). Then I - 3ln(sqrt3) = I - (3/2)ln3 = 2 - sqrt2 - ln(1+sqrt2). Hence option B.

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
2 April, Evening Shift
Subject
Mathematics
Chapter
Definite Integration
Topic
Evaluation of Definite Integrals

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