JEE Main 2025 Jan 23 Shift 1, Mathematics Q1: Some Properties of Definite Integrals

JEE Main2025Jan 23, Shift 1Mathematics

The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e ^{((l o g_{e} x)^{2} + 1)^{- 1}}}{e ^{((l o g_{e} x)^{2} + 1)^{- 1}} + e ^{((6 - l o g_{e} x)^{2} + 1)^{- 1}}}) d x$ is

A
$lo g_{e} 2$
B
$2$
C
$1$
D
$e^{2}$

Solution

Substitute $t = ln x$ , so $\frac{d x}{x} = d t$ and the limits become $t = 2$ to $t = 4$ .

$I = \int_{2}^{4} \frac{e ^{1/ (1 + t^{2})}}{e ^{1/ (1 + t^{2})} + e ^{1/ (1 + (6 - t)^{2})}} d t$

Now apply the King property $\int_{a}^{b} f (t) d t = \int_{a}^{b} f (a + b - t) d t$ with $a + b = 6$ . Replacing $t \to 6 - t$ swaps the two exponentials:

$I = \int_{2}^{4} \frac{e ^{1/ (1 + (6 - t)^{2})}}{e ^{1/ (1 + (6 - t)^{2})} + e ^{1/ (1 + t^{2})}} d t$

Adding the two forms, the integrand sums to 1, giving $2 I = \int_{2}^{4} d t = 2$ .

$∴ I = 1$ .

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JEE Main 2025 Jan 23 Shift 1, Mathematics Q1: Some Properties of Definite Integrals

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