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 evaluate the integral

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

, we follow these steps.

Step 1: Substitution and Transformation

Let \( t = \log_e x \). This implies \( dt = \frac{1}{x} dx \). The integration limits transform as follows: 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: Exploiting Symmetry

The integrand exhibits symmetry around \( t = 3 \). Let \( t = 3 + u \). The new integration limits are: when \( t = 2 \), \( u = -1 \); when \( t = 4 \), \( u = 1 \).

The integral transforms to:

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

The integrand is symmetric about \( u = 0 \). This means the integral from \(-1\) to \(0\) equals the integral from \(0\) to \(1\), and the total integral is twice the integral from \(0\) to \(1\).

Step 3: Final Calculation

Due to the symmetry around \( u=0 \), the integrand simplifies to \(1/2\) over the interval \([-1, 1]\). This occurs because the terms in the numerator and denominator become equal due to the reciprocal and logarithmic symmetry.

The integral evaluates to:

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

Conclusion

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