JEE Main 2025 — 23 January, Morning Shift — Question 1

$\text{Let\; } t = \log_e x \Rightarrow dt = \frac{dx}{x}$

$\text{When\; } x = e^2, \; t = 2 \quad \text{and\; when\; } x = e^4, \; t = 4$

$\therefore I = \int_{2}^{4} \frac{e^{\frac{1}{t^2 + 1}}}{e^{\frac{1}{t^2 + 1}} + e^{\frac{1}{(6 - t)^2 + 1}}} \, dt$

$\text{Let\; } f(t) = \frac{e^{\frac{1}{t^2 + 1}}}{e^{\frac{1}{t^2 + 1}} + e^{\frac{1}{(6 - t)^2 + 1}}}$

$\text{Then\; } f(6 - t) = \frac{e^{\frac{1}{(6 - t)^2 + 1}}}{e^{\frac{1}{(6 - t)^2 + 1}} + e^{\frac{1}{t^2 + 1}}}$

$\Rightarrow f(t) + f(6 - t) = 1$

$\therefore 2I = \int_{2}^{4} \big[f(t) + f(6 - t)\big] \, dt = \int_{2}^{4} 1 \, dt = 2$

$\Rightarrow I = 1$

$\boxed{1}$