Question

Evaluate \( \int_{0}^{1} \frac{e^{-x}}{1 + e^x} \, dx \).

Hint:

Use substitution and partial fraction decomposition to simplify integrals involving rational expressions.

Answer 1

Step 1: Simplify the integral: Let \( I = \int_{0}^{1} \frac{e^{-x}}{1 + e^x} \, dx \). Substitute \( u = e^x \), so \( du = e^x dx \) and the limits become \( u = 1 \) (at \( x = 0 \)) and \( u = e \) (at \( x = 1 \)). 
Step 2: Rewrite the integral: \[ I = \int_{1}^{e} \frac{1}{1 + u} \cdot \frac{1}{u} \, du = \int_{1}^{e} \frac{1}{u(1 + u)} \, du. \] 
Step 3: Use partial fraction decomposition: \[ \frac{1}{u(1 + u)} = \frac{A}{u} + \frac{B}{1 + u}. \] Solve for \( A \) and \( B \): \[ 1 = A(1 + u) + Bu \quad \Rightarrow \quad A = 1, \, B = -1. \] Thus: \[ \frac{1}{u(1 + u)} = \frac{1}{u} - \frac{1}{1 + u}. \] 
Step 4: Substitute back and integrate: \[ I = \int_{1}^{e} \left( \frac{1}{u} - \frac{1}{1 + u} \right) \, du = \left[ \ln{u} - \ln{(1 + u)} \right]_{1}^{e}. \] Simplify: \[ I = \left( \ln{e} - \ln{(1 + e)} \right) - \left( \ln{1} - \ln{2} \right) = \left( 1 - \ln{(1 + e)} \right) - \left( 0 - \ln{2} \right). \] \[ I = 1 - \ln{(1 + e)} + \ln{2}. \]