JEE MainMathematics

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Definite Integral Involving King's Property After Substitution

The value of integral from e^2 to e^4 of 1/x ((e^(((log_e x)^2+1)^(-1)))/(e^(((log_e x)^2+1)^(-1)) + e^(((6-log_e x)^2+1)^(-1)))) dx is

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Step-by-step solutionView

Correct answer

Using the substitution $t = \log_{e} x$ transforms the integral into $\int_{2}^{4} \frac{f (t)}{f (t) + f (6 - t)} d t$ , 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 $2 I = b - a$ . Remember that adding the two symmetric forms yields $2 I = \int_{a}^{b} 1 d t = b - a$ , 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 $1 / x d x$ as $d t$ under the substitution $t = \log_{e} x$ rather than separating the integrand into unrelated products.

C · 1

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 $d x / x$ 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

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

02Approach

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

03Execute

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

04Execute

Using King's property, replace $t$ by $2 + 4 - t = 6 - t$ : $I = \int_{2}^{4} \frac{e^{\frac{1}{(6 - t)^{2} + 1}}}{e^{\frac{1}{(6 - t)^{2} + 1}} + e^{\frac{1}{t^{2} + 1}}} d t$ . Adding the two expressions for $I$ : $2 I = \int_{2}^{4} (\frac{e^{\frac{1}{t^{2} + 1}} + e^{\frac{1}{(6 - t)^{2} + 1}}}{e^{\frac{1}{t^{2} + 1}} + e^{\frac{1}{(6 - t)^{2} + 1}}}) d t = \int_{2}^{4} 1 d t = 4 - 2 = 2$ . 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 $I = \int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e^{((\log_{e} x)^{2} + 1)^{- 1}}}{e^{((\log_{e} x)^{2} + 1)^{- 1}} + e^{((6 - \log_{e} x)^{2} + 1)^{- 1}}}) d x$ ?
Let $t = \log_{e} x$ , so $d t = \frac{1}{x} d x$ , 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?
$I = \int_{2}^{4} \frac{f (t)}{f (t) + f (6 - t)} d t$ , where $f (t) = e^{(t^{2} + 1)^{- 1}}$
Applying King's property $\int_{a}^{b} g (t) d t = \int_{a}^{b} g (a + b - t) d t$ 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 $2 I = \int_{2}^{4} 1 d t = 4 - 2 = 2$ , so $I = 1$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $\int_{2}^{4} \frac{f (t)}{f (t) + f (6 - t)} d t$ , 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 $2 I = b - a$ . Remember that adding the two symmetric forms yields $2 I = \int_{a}^{b} 1 d t = b - a$ , 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 $1 / x d x$ as $d t$ under the substitution $t = \log_{e} x$ rather than separating the integrand into unrelated products.
C · correct: 1 - 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 $d x / x$ 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: $I = \int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e^{\frac{1}{(\ln x)^{2} + 1}}}{e^{\frac{1}{(\ln x)^{2} + 1}} + e^{\frac{1}{(6 - \ln x)^{2} + 1}}}) d x$
Approach: Substitute $t = \ln x$ so that $d t = \frac{1}{x} d x$ . Then apply King's Property $\int_{a}^{b} f (t) d t = \int_{a}^{b} f (a + b - t) d t$ .
Execute: Let $t = \ln x$ , then $d t = \frac{d x}{x}$ . For $x = e^{2}$ , $t = 2$ ; for $x = e^{4}$ , $t = 4$ . Thus, $I = \int_{2}^{4} \frac{e^{\frac{1}{t^{2} + 1}}}{e^{\frac{1}{t^{2} + 1}} + e^{\frac{1}{(6 - t)^{2} + 1}}} d t$ .
Execute: Using King's property, replace $t$ by $2 + 4 - t = 6 - t$ : $I = \int_{2}^{4} \frac{e^{\frac{1}{(6 - t)^{2} + 1}}}{e^{\frac{1}{(6 - t)^{2} + 1}} + e^{\frac{1}{t^{2} + 1}}} d t$ . Adding the two expressions for $d t = \frac{1}{x} d x$ : $2 I = \int_{2}^{4} (\frac{e^{\frac{1}{t^{2} + 1}} + e^{\frac{1}{(6 - t)^{2} + 1}}}{e^{\frac{1}{t^{2} + 1}} + e^{\frac{1}{(6 - t)^{2} + 1}}}) d t = \int_{2}^{4} 1 d t = 4 - 2 = 2$ . 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 $\frac{4 - 2}{2} = 1$ .

Definite Integral Involving King's Property After Substitution

What feels right?

Using the substitution $t = \log_{e} x$ transforms the integral into $\int_{2}^{4} \frac{f (t)}{f (t) + f (6 - t)} d t$ , 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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