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Integral Calculus: JEE Main Mathematics Question with Solution

The value of

\int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{((\log_e x)^2+1)^{-1}}}{e^{((\log_e x)^2+1)^{-1}} + e^{((6-\log_e x)^2+1)^{-1}}} \right) dx

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What feels right?

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A2B

\log_e 2

C1D

e^2

Step-by-step solutionView

Correct answer

Using the substitution $t = \log_e x$ transforms the integral into $\int_2^4 \frac{f(t)}{f(t)+f(6-t)} dt$ , which evaluates to $\frac{4-2}{2} = 1$ by King's property.

Option analysis

Why each option works or fails

A · 2

The student computed the length of the interval of integration $4 - 2 = 2$ , forgetting to divide by $2$ when solving $2I = b - a$ . Remember that adding the two symmetric forms yields $2I = \int_a^b 1\, dt = b - a$ , so $I = \frac{b - a}{2}$ .

B ·

\log_e 2

The student incorrectly assumed the $1/x$ factor integrates directly to $\log_e(x)$ evaluated from $e^2$ to $e^4$ , giving $\log_e(4/2) = \log_e 2$ , neglecting the integrand's main fraction. Use $1/x\,dx$ as $dt$ under the substitution $t = \log_e x$ rather than separating the integrand into unrelated products.

C · 1

None. The student correctly applied substitution and King's property to obtain $I = (4 - 2)/2 = 1$ . This is the correct answer.

D ·

e^2

The student attempted to use the symmetry property directly on the original variable $x$ without transforming limits or accounting for the $dx/x$ differential, arriving at $\frac{e^4 - e^2}{...}$ and confusing it with $e^2$ . Always perform the substitution $t = \log_e x$ first to linearize the limits and simplify the differential before applying King's property.

Reviewed route

Solution

StepWorking

01given

$I = \int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{\frac{1}{(\ln x)^2 + 1}}}{e^{\frac{1}{(\ln x)^2 + 1}} + e^{\frac{1}{(6 - \ln x)^2 + 1}}} \right) dx$

02approach

Substitute $t = \ln x$ so that $dt = \frac{1}{x} dx$ . Then apply King's Property $\int_a^b f(t) dt = \int_a^b f(a+b-t) dt$ .

03execute

Let $t = \ln x$ , then $dt = \frac{dx}{x}$ . For $x = e^2$ , $t = 2$ ; for $x = e^4$ , $t = 4$ . Thus, $I = \int_{2}^{4} \frac{e^{\frac{1}{t^2 + 1}}}{e^{\frac{1}{t^2 + 1}} + e^{\frac{1}{(6 - t)^2 + 1}}} dt$ .

04execute

Using King's property, replace $t$ by $2 + 4 - t = 6 - t$ : $I = \int_{2}^{4} \frac{e^{\frac{1}{(6 - t)^2 + 1}}}{e^{\frac{1}{(6 - t)^2 + 1}} + e^{\frac{1}{t^2 + 1}}} dt$ . Adding the two expressions for $I$ : $2I = \int_{2}^{4} \left( \frac{e^{\frac{1}{t^2 + 1}} + e^{\frac{1}{(6 - t)^2 + 1}}}{e^{\frac{1}{t^2 + 1}} + e^{\frac{1}{(6 - t)^2 + 1}}} \right) dt = \int_{2}^{4} 1 \, dt = 4 - 2 = 2$ . Therefore, $I = 1$ .

✓verify

The integrand in $t$ is of the form $\frac{g(t)}{g(t) + g(6-t)}$ , which has symmetry about the midpoint $t=3$ . The integral over $[2, 4]$ is identically $\frac{4 - 2}{2} = 1$ .

Hints that build this answer step by step

What is the most effective initial substitution to simplify the integral $I = \int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{((\log_e x)^2+1)^{-1}}}{e^{((\log_e x)^2+1)^{-1}} + e^{((6-\log_e x)^2+1)^{-1}}} \right) dx$ ?
Let $t = \log_e x$ , so $dt = \frac{1}{x} dx$ , transforming the limits from $x \in [e^2, e^4]$ to $t \in [2, 4]$ .
After substituting $t = \log_e x$ , what form does the integral take?
$I = \int_2^4 \frac{f(t)}{f(t) + f(6-t)} dt$ , where $f(t) = e^{(t^2+1)^{-1}}$
Applying King's property $\int_a^b g(t)\, dt = \int_a^b g(a+b-t)\, dt$ with $a=2$ and $b=4$ , what is $a+b-t$ and what is the final value of $I$ ?
$a+b-t = 6-t$ , leading to $2I = \int_2^4 1\, dt = 4 - 2 = 2$ , so $I = 1$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why do we let t = ln x first?

Because the derivative of ln x is 1/x, which directly cancels the 1/x factor in dx, simplifying the integrand to a standard King's property form.

Answer

Using the substitution $t = \log_e x$ transforms the integral into $\int_2^4 \frac{f(t)}{f(t)+f(6-t)} dt$ , which evaluates to $\frac{4-2}{2} = 1$ by King's property.

Why each option works or fails

A: 2 - The student computed the length of the interval of integration $4 - 2 = 2$ , forgetting to divide by $2$ when solving $2I = b - a$ . Remember that adding the two symmetric forms yields $2I = \int_a^b 1\, dt = b - a$ , so $I = \frac{b - a}{2}$ .
B: $\log_e 2$ - The student incorrectly assumed the $1/x$ factor integrates directly to $\log_e(x)$ evaluated from $e^2$ to $e^4$ , giving $\log_e(4/2) = \log_e 2$ , neglecting the integrand's main fraction. Use $1/x\,dx$ as $dt$ under the substitution $t = \log_e x$ rather than separating the integrand into unrelated products.
C · correct: 1 - None. The student correctly applied substitution and King's property to obtain $I = (4 - 2)/2 = 1$ . This is the correct answer.
D: $e^2$ - The student attempted to use the symmetry property directly on the original variable $x$ without transforming limits or accounting for the $dx/x$ differential, arriving at $\frac{e^4 - e^2}{...}$ and confusing it with $e^2$ . Always perform the substitution $t = \log_e x$ first to linearize the limits and simplify the differential before applying King's property.

Step-by-step solution

given: $I = \int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{\frac{1}{(\ln x)^2 + 1}}}{e^{\frac{1}{(\ln x)^2 + 1}} + e^{\frac{1}{(6 - \ln x)^2 + 1}}} \right) dx$
approach: Substitute $t = \ln x$ so that $dt = \frac{1}{x} dx$ . Then apply King's Property $\int_a^b f(t) dt = \int_a^b f(a+b-t) dt$ .
execute: Let $t = \ln x$ , then $dt = \frac{dx}{x}$ . For $x = e^2$ , $t = 2$ ; for $x = e^4$ , $t = 4$ . Thus, $I = \int_{2}^{4} \frac{e^{\frac{1}{t^2 + 1}}}{e^{\frac{1}{t^2 + 1}} + e^{\frac{1}{(6 - t)^2 + 1}}} dt$ .
execute: Using King's property, replace $t$ by $2 + 4 - t = 6 - t$ : $I = \int_{2}^{4} \frac{e^{\frac{1}{(6 - t)^2 + 1}}}{e^{\frac{1}{(6 - t)^2 + 1}} + e^{\frac{1}{t^2 + 1}}} dt$ . Adding the two expressions for $I$ : $2I = \int_{2}^{4} \left( \frac{e^{\frac{1}{t^2 + 1}} + e^{\frac{1}{(6 - t)^2 + 1}}}{e^{\frac{1}{t^2 + 1}} + e^{\frac{1}{(6 - t)^2 + 1}}} \right) dt = \int_{2}^{4} 1 \, dt = 4 - 2 = 2$ . Therefore, $I = 1$ .
verify: The integrand in $t$ is of the form $\frac{g(t)}{g(t) + g(6-t)}$ , which has symmetry about the midpoint $t=3$ . The integral over $[2, 4]$ is identically $t = \ln x$ .

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

Using the substitution $t = \log_e x$ transforms the integral into $\int_2^4 \frac{f(t)}{f(t)+f(6-t)} dt$ , which evaluates to $\frac{4-2}{2} = 1$ by King's property.

Why each option works or fails

Solution

01given

02approach

03execute

04execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

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Students also ask

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The value of $\int_{0}^{20\pi} (\sin^4 x + \cos^4 x)dx$ is equal to…