Logo
Jeeify
MCQMediumJEE 2025Definite Integrals

JEE Mathematics 2025 Question with Solution

Scanned mathematics section image containing the definite integral question and its four answer options from JEE Main paper.

MATHEMATICS SECTION-A

The value of ∫e2e41x(e((log⁡ex)2+1)−1e((log⁡ex)2+1)−1+e((6−log⁡ex)2+1)−1)dx\int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{\left( (\log_e x)^2 +1 \right)^{-1}}}{e^{\left( (\log_e x)^2 +1 \right)^{-1}} + e^{\left( (6-\log_e x)^2 +1 \right)^{-1}}} \right) dx is:

  • A

    log⁡2\log 2

  • B

    22

  • C

    11

  • D

    e2e^2

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

I=∫e2e41x(e((log⁡ex)2+1)−1e((log⁡ex)2+1)−1+e((6−log⁡ex)2+1)−1)dxI=\int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{\left( (\log_e x)^2 +1 \right)^{-1}}}{e^{\left( (\log_e x)^2 +1 \right)^{-1}} + e^{\left( (6-\log_e x)^2 +1 \right)^{-1}}} \right) dx

Find: The value of the integral.

Let

t=log⁡ext=\log_e x

Then

dt=1xdxdt=\frac{1}{x}dx

When x=e2x=e^2, t=2t=2 and when x=e4x=e^4, t=4t=4.

So the integral becomes

I=∫24e(t2+1)−1e(t2+1)−1+e((6−t)2+1)−1 dtI=\int_{2}^{4} \frac{e^{\left( t^2 + 1 \right)^{-1}}}{e^{\left( t^2 + 1 \right)^{-1}} + e^{\left( (6-t)^2 + 1 \right)^{-1}}} \, dt

Now use the symmetry substitution

t↦6−tt \mapsto 6-t

Then the same integral can be written as

I=∫24e((6−t)2+1)−1e((6−t)2+1)−1+e(t2+1)−1 dtI=\int_{2}^{4} \frac{e^{\left( (6-t)^2 + 1 \right)^{-1}}}{e^{\left( (6-t)^2 + 1 \right)^{-1}} + e^{\left( t^2 + 1 \right)^{-1}}} \, dt

Adding the two expressions,

2I=∫24(e(t2+1)−1e(t2+1)−1+e((6−t)2+1)−1+e((6−t)2+1)−1e((6−t)2+1)−1+e(t2+1)−1)dt=∫241 dt=4−2=2\begin{aligned} 2I &= \int_{2}^{4} \left( \frac{e^{\left( t^2 + 1 \right)^{-1}}}{e^{\left( t^2 + 1 \right)^{-1}} + e^{\left( (6-t)^2 + 1 \right)^{-1}}} + \frac{e^{\left( (6-t)^2 + 1 \right)^{-1}}}{e^{\left( (6-t)^2 + 1 \right)^{-1}} + e^{\left( t^2 + 1 \right)^{-1}}} \right) dt \\ &= \int_{2}^{4} 1 \, dt \\ &= 4-2 = 2 \end{aligned}

Hence,

I=1I=1

Therefore, the correct option is C.

Symmetry Trick

Given: The integrand has the form

f(t)=A(t)A(t)+A(6−t)f(t)=\frac{A(t)}{A(t)+A(6-t)}

after putting t=log⁡ext=\log_e x.

Find: The value of the definite integral.

With

t=log⁡ex,dt=1xdxt=\log_e x, \qquad dt=\frac{1}{x}dx

we get

I=∫24f(t) dtI=\int_2^4 f(t) \, dt

where

f(t)=e(t2+1)−1e(t2+1)−1+e((6−t)2+1)−1f(t)=\frac{e^{\left( t^2 + 1 \right)^{-1}}}{e^{\left( t^2 + 1 \right)^{-1}}+e^{\left( (6-t)^2 + 1 \right)^{-1}}}

Now observe that

f(6−t)=e((6−t)2+1)−1e((6−t)2+1)−1+e(t2+1)−1f(6-t)=\frac{e^{\left( (6-t)^2 + 1 \right)^{-1}}}{e^{\left( (6-t)^2 + 1 \right)^{-1}}+e^{\left( t^2 + 1 \right)^{-1}}}

So,

f(t)+f(6−t)=1f(t)+f(6-t)=1

For an integral over [2,4][2,4], this symmetry gives

2I=∫24(f(t)+f(6−t)) dt=∫241 dt=22I=\int_2^4 \big(f(t)+f(6-t)\big) \, dt=\int_2^4 1 \, dt=2

Thus,

I=1I=1

Therefore, the value of the integral is 11.

Common mistakes

  • Using t=log⁡ext=\log_e x but forgetting that dt=1xdxdt=\frac{1}{x}dx. This is wrong because the factor 1x\frac{1}{x} is essential for the substitution. Replace the entire term 1xdx\frac{1}{x}dx by dtdt and also change the limits.

  • Treating the integrand as constant 12\frac{1}{2} without first establishing the symmetry. This is wrong because the numerator and denominator are not individually equal for every tt. First show that f(t)+f(6−t)=1f(t)+f(6-t)=1, then use definite-integral symmetry.

  • Not changing the limits after substitution and continuing with xx-limits in the tt-integral. This mixes two variables incorrectly. After t=log⁡ext=\log_e x, the limits must become 22 and 44.

Practice more Definite Integrals questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Definite Integrals

Related questions