Mathematics · Definite Integration

JEE Main 2026 — 8 April, Evening Shift — Question 33

The value of the integral ∫02x(x2+x+1)(x+1)(x4+x2+1)dx\int_{0}^{2}\frac{\sqrt{x(x^{2}+x+1)}}{(\sqrt{x+1})(\sqrt{x^{4}+x^{2}+1})}dx is equal to :

  1. Option A:

    13log⁡e(3−22)\frac{1}{3}\log_{e}(3-2\sqrt{2})

  2. Option B:

    23log⁡e(4+2)\frac{2}{3}\log_{e}(4+\sqrt{2})

  3. Option C:

    23log⁡e(3+22)\frac{2}{3}\log_{e}(3+2\sqrt{2})

    Correct
  4. Option D:

    13log⁡e(1+62)\frac{1}{3}\log_{e}(1+6\sqrt{2})

Answer: C

Step-by-step solution

I=∫02x(x2+x+1)(x+1)(x2−x+1)(x2+x+1)dxI=\int_{0}^{2} \sqrt{\frac{x\left(x^{2}+x+1\right)}{(x+1)\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)} d x} =∫02xx3+1dx=\int_{0}^{2} \sqrt{\frac{\mathrm{x}}{\mathrm{x}^{3}+1}} \mathrm{dx} (x3/2=t⇒32x1/2dx=dt)\left(\mathrm{x}^{3 / 2}=\mathrm{t} \Rightarrow \frac{3}{2} \mathrm{x}^{1 / 2} \mathrm{dx}=\mathrm{dt}\right) =23∫023/2dtt2+1=\frac{2}{3} \int_{0}^{2^{3 / 2}} \frac{\mathrm{dt}}{\sqrt{\mathrm{t}^{2}+1}} =23(ln⁡(t+t2+1))023/2=\frac{2}{3}\left(\ln \left(\mathrm{t}+\sqrt{\mathrm{t}^{2}+1}\right)\right)_{0}^{2^{3 / 2}} =23ln⁡(23/2+3)=\frac{2}{3} \ln \left(2^{3 / 2}+3\right)

23log⁡e(3+22)\frac{2}{3}\log_{e}(3+2\sqrt{2})

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Exam
JEE Main 2026
Subject
Mathematics
Chapter
Definite Integration
Topic
Evaluation of Definite Integrals
The value of the integral int 0 2 frac sqrt x(x 2 +x+1) (√(x+1))(sqrt… | JEE Main 2026 PYQ with Solution · DhiX AI