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

Step 1: Substituting \( \ln x = t \), we get: \[ dx = x dt \Rightarrow \frac{dx}{x} = dt \] Thus, the given integral transforms as: \[ I = \int_{2}^{4} \frac{e^{1+t^2}}{e^{1+t^2} + e^{1+(6-t)^2}} dt. \] Now, using the property: \[ I = \int_{2}^{4} \frac{e^{1+(6-t)^2}}{e^{1+(6-t)^2} + e^{1+t^2}} dt. \] Step 2: Adding both integrals: \[ 2I = \int_{2}^{4} dt. \] Evaluating, \[ 2I = (t) \Big|_{2}^{4} = 4 - 2 = 2. \] Thus, \[ I = 1. \]