JEE Main 2026 — Mathematics Calculus

Let $\displaystyle I = \displaystyle\int_{0}^{2} \dfrac{\sqrt{x(x^2 + x + 1)}}{\sqrt{x + 1}\,\sqrt{x^4 + x^2 + 1}}\, dx$I=∫02​x+1​x4+x2+1​x(x2+x+1)​​dx.

Using the identity $x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$x4+x2+1=(x2+x+1)(x2−x+1):

$$I = \displaystyle\int_{0}^{2} \dfrac{\sqrt{x}\,\sqrt{x^2 + x + 1}}{\sqrt{x + 1}\,\sqrt{(x^2 + x + 1)(x^2 - x + 1)}}\, dx$$I=∫02​x+1​(x2+x+1)(x2−x+1)​x​x2+x+1​​dx

Cancelling $\sqrt{x^2 + x + 1}$x2+x+1​:

$$I = \displaystyle\int_{0}^{2} \dfrac{\sqrt{x}}{\sqrt{(x + 1)(x^2 - x + 1)}}\, dx$$I=∫02​(x+1)(x2−x+1)​x​​dx

Using $(x + 1)(x^2 - x + 1) = x^3 + 1$(x+1)(x2−x+1)=x3+1:

$$I = \displaystyle\int_{0}^{2} \dfrac{\sqrt{x}}{\sqrt{x^3 + 1}}\, dx$$I=∫02​x3+1​x​​dx

Substitution:

Let $t = x^{3/2}$t=x3/2, so $dt = \dfrac{3}{2}x^{1/2}\, dx \Rightarrow \sqrt{x}\, dx = \dfrac{2}{3}\, dt$dt=23​x1/2dx⇒x​dx=32​dt.

Also, $t^2 = x^3$t2=x3, so $x^3 + 1 = t^2 + 1$x3+1=t2+1.

Changing limits:

$$x = 0 \Rightarrow t = 0$$x=0⇒t=0

$$x = 2 \Rightarrow t = 2^{3/2} = 2\sqrt{2}$$x=2⇒t=23/2=22​

Therefore:

$$I = \dfrac{2}{3}\displaystyle\int_{0}^{2\sqrt{2}} \dfrac{dt}{\sqrt{t^2 + 1}}$$I=32​∫022​​t2+1​dt​

Using $\displaystyle \displaystyle\int \dfrac{dt}{\sqrt{t^2 + a^2}} = \log_e\left|t + \sqrt{t^2 + a^2}\right| + C$∫t2+a2​dt​=loge​​t+t2+a2​​+C:

$$I = \dfrac{2}{3}\left[\log_e\left(t + \sqrt{t^2 + 1}\right)\right]_{0}^{2\sqrt{2}}$$I=32​[loge​(t+t2+1​)]022​​

$$I = \dfrac{2}{3}\left[\log_e\left(2\sqrt{2} + \sqrt{8 + 1}\right) - \log_e(1)\right]$$I=32​[loge​(22​+8+1​)−loge​(1)]

$$I = \dfrac{2}{3}\log_e\left(2\sqrt{2} + 3\right)$$I=32​loge​(22​+3)

$$I = \dfrac{2}{3}\log_e\left(3 + 2\sqrt{2}\right)$$I=32​loge​(3+22​)

Hence, the correct option is $(3)\ \dfrac{2}{3}\log_e\left(3 + 2\sqrt{2}\right)$(3) 32​loge​(3+22​).