JEE MAIN 2026 · Integral Calculus

Simplify. $x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$, so the factor $\sqrt{x^2 + x + 1}$ cancels and

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

Substitute $u = x^{3/2}$. Then $du = \frac32\sqrt x\,dx$, and $x^3 = u^2$; the limits become $0$ and $2\sqrt2$:

$$\int_0^2 \frac{\sqrt x\,dx}{\sqrt{x^3 + 1}} = \frac23 \int_0^{2\sqrt2} \frac{du}{\sqrt{u^2 + 1}} = \frac23 \Big[\ln\big(u + \sqrt{u^2 + 1}\big)\Big]_0^{2\sqrt2} = \frac23 \ln\big(2\sqrt2 + 3\big).$$

Answer: $\frac23 \log_e(3 + 2\sqrt2)$.