The value of the integral $\int_{0}^{2} \frac{\sqrt{x\left(x^{2}+x+1\right)}}{(\sqrt{x+1})\left(\sqrt{x^{4}+x^{2}+1}\right)} d x$ is equal to :

(C) $\frac{2}{3} \log _{e}(3+2 \sqrt{2})$

(A) $\frac{1}{3} \log _{e}(3-2 \sqrt{2})$

(B) $\frac{2}{3} \log _{e}(4+\sqrt{2})$

(D) $\frac{1}{3} \log _{e}(1+6 \sqrt{2})$

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Definite Integrals

The value of the integral

\int_{0}^{2} \frac{x ( x ^{2} + x + 1 )}{( x + 1 ) ( x ^{4} + x ^{2} + 1 )} d x

is equal to :

Answer: (C)

Factor

x^{4} + x^{2} + 1 = (x^{2} + x + 1) (x^{2} - x + 1)

, so the integrand simplifies:

\frac{x ( x ^{2} + x + 1 )}{( x + 1 ) ( x ^{2} + x + 1 ) ( x ^{2} - x + 1 )} = \frac{x}{( x + 1 ) ( x ^{2} - x + 1 )} = \frac{x}{x ^{3} + 1} .

Put

t = x^{3/2}

, so

d t = \frac{3}{2} x^{1/2} d x

and the integral becomes

\frac{2}{3} \int_{0}^{22} \frac{d t}{t ^{2} + 1} = \frac{2}{3} [ln (t + t^{2} + 1)]_{0}^{22} = \frac{2}{3} ln (22 + 3) = \frac{2}{3} lo g_{e} (3 + 22) .

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option