JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 16

Simplify. $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ and $(x+1)(x^2-x+1)=x^3+1$, so the integrand is
$$\sqrt{\frac x{x^3+1}}.$$

Substitute $t=x^{3/2}$, $dt=\tfrac32\sqrt x\,dx$:
$$\frac23\int_0^{2\sqrt2}\frac{dt}{\sqrt{1+t^2}}=\frac23\ln\left(2\sqrt2+3\right).$$

(C)