JEE Main 2026 Apr 8 Shift 2, Mathematics Q16: Methods of Integration

Simplify the integrand. Note $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$, and $x^3+1=(x+1)(x^2-x+1)$.

$\frac{x(x^2+x+1)}{(x+1)(x^4+x^2+1)}=\frac{x(x^2+x+1)}{(x+1)(x^2+x+1)(x^2-x+1)}=\frac{x}{(x+1)(x^2-x+1)}=\frac{x}{x^3+1}$

So $I={\int }_0^2\sqrt{\frac{x}{x^3+1}}dx$.

Substitute $t=x^{3/2}$, so $dt=\frac{3}{2}{x}^{1/2}dx$, hence $\sqrt{x}dx=\frac{2}{3}dt$, and $x^3=t^2$.

$I={\int }_0^{2\sqrt{2}}\frac{1}{\sqrt{t^2+1}}\cdot \frac{2}{3}dt=\frac{2}{3}\left[\ln \right(t+\sqrt{t^2+1}){]}_0^{2\sqrt{2}}$

$=\frac{2}{3}\ln \left(2\sqrt{2}+\sqrt{8+1}\right)=\frac{2}{3}\ln (2\sqrt{2}+3)=\frac{2}{3}{\log }_e(3+2\sqrt{2})$.