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JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Definite Integrals

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}}}) dx

is

(A) $lo g_{e} 2$
(B)2
(C)1✓ Correct
(D) $e^{2}$

Answer: (C)

Solution

Substitute

t = lo g_{e} x

, so

d t = \frac{d x}{x}

; when

x = e^{2}

,

t = 2

, and when

x = e^{4}

,

t = 4

:

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

Apply

\int_{a}^{b} f (t) d t = \int_{a}^{b} f (a + b - t) d t

with

a + b = 6

, replacing

t

by

6 - t

:

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

Adding the two expressions for

I

, the fractions combine to

1

at every point:

2 I = \int_{2}^{4} 1 d t = 2 ⟹ I = 1

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option

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