Question

The value of \[ \int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{\left( (\log_e x)^2 +1 \right)^{-1}}}{e^{\left( (\log_e x)^2 +1 \right)^{-1}} + e^{\left( (6-\log_e x)^2 +1 \right)^{-1}}} \right) dx \] is:

• A. \( \log 2 \)
• B. \( 2 \)
• C. \( 1 \)
• D. \( e^2 \)

Hint:

Use substitution techniques effectively in definite integrals to simplify the given expression. Symmetric properties in definite integrals can often reduce complex terms.

Answer 1

The Correct Option is C

To solve the integral 

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

, let's proceed step by step.

Step 1: Simplification of the integral expression 

First, denote \(t = \log_e x\). Then, the differential \(dt = \frac{1}{x} dx\), which implies that the limits of the integration also change when substituting \( x = e^t \): - When \( x = e^2 \), \( t = 2 \). - When \( x = e^4 \), \( t = 4 \).

The integral becomes: 

\[\int_{2}^{4} \frac{e^{\left( t^2 + 1 \right)^{-1}}}{e^{\left( t^2 + 1 \right)^{-1}} + e^{\left( (6-t)^2 + 1 \right)^{-1}}} \, dt\]

Step 2: Symmetry Analysis

Observe that the expression inside the integrand exhibits symmetry around \( t = 3 \).

If we substitute \( t = 3 + u \), then the range changes to: - When \( t = 2 \), \( u = -1 \). - When \( t = 4 \), \( u = 1 \).

Thus, the integral becomes: 

\[\int_{-1}^{1} \frac{e^{\left( (3+u)^2 + 1 \right)^{-1}}}{e^{\left( (3+u)^2 + 1 \right)^{-1}} + e^{\left( (3-u)^2 + 1 \right)^{-1}}} \, du\]

The function is symmetric about \( u = 0 \), which implies:

• The integral from \(-1\) to \(0\) is equal to the integral from \(0\) to \(1\).
• This simplifies to twice the integral from \(0\) to \(1\).

Step 3: Calculation of the simplified integral

By the property of symmetry, the function with respect to \( u =0 \) yields:

The integrand becomes \(1/2\) over the interval \([-1, 1]\), as both terms are equal due to their reciprocal and logarithmic symmetry.

So the integral simplifies to: 

\[\int_{-1}^{1} \frac{1}{2} \, du = \frac{1}{2} \times (1 - (-1)) = 1\]

Conclusion

The value of the integral is \(1\). Therefore, the correct answer is \(\boxed{1}\).

Answer 2

Step 1: Variable Substitution
Let \(\ln x = t\), which gives: \[ \frac{dx}{x} = dt \] The integral transforms to: \[ I = \int_{2}^{4} \frac{e^{1+t^2}}{e^{1+t^2} + e^{1+(6-t)^2}} dt \]

Step 2: Symmetry Property Application
Using the property of definite integrals, we can write: \[ I = \int_{2}^{4} \frac{e^{1+(6-t)^2}}{e^{1+(6-t)^2} + e^{1+t^2}} dt \]

Step 3: Combining Integrals
Adding both expressions for \(I\): \[ 2I = \int_{2}^{4} \left( \frac{e^{1+t^2} + e^{1+(6-t)^2}}{e^{1+t^2} + e^{1+(6-t)^2}} \right) dt = \int_{2}^{4} 1 \, dt \]

Step 4: Evaluation
Calculating the integral: \[ 2I = (t) \Big|_{2}^{4} = 4 - 2 = 2 \] Therefore: \[ I = 1 \]

Final Answer:
The value of the integral is \(1\).