Step 1: Define \( I = \int_{0}^{1} \frac{e^{-x}}{1 + e^x} \, dx \). Perform the substitution \( u = e^x \), which implies \( du = e^x dx \). The integration limits change to \( u = 1 \) at \( x = 0 \) and \( u = e \) at \( x = 1 \).
Step 2: Rewrite the integral with the substitution: \[ I = \int_{1}^{e} \frac{1}{1 + u} \cdot \frac{1}{u} \, du = \int_{1}^{e} \frac{1}{u(1 + u)} \, du. \]
Step 3: Decompose the integrand using partial fractions: \[ \frac{1}{u(1 + u)} = \frac{A}{u} + \frac{B}{1 + u}. \] Solving for constants yields \( A = 1 \) and \( B = -1 \). Therefore: \[ \frac{1}{u(1 + u)} = \frac{1}{u} - \frac{1}{1 + u}. \]
Step 4: Substitute the partial fraction decomposition back into the integral and evaluate: \[ I = \int_{1}^{e} \left( \frac{1}{u} - \frac{1}{1 + u} \right) \, du = \left[ \ln{u} - \ln{(1 + u)} \right]_{1}^{e}. \] Applying the limits of integration: \[ I = \left( \ln{e} - \ln{(1 + e)} \right) - \left( \ln{1} - \ln{2} \right) = \left( 1 - \ln{(1 + e)} \right) - \left( 0 - \ln{2} \right). \] Simplifying the expression gives: \[ I = 1 - \ln{(1 + e)} + \ln{2}. \]