\( \int \frac { 1 } { 1 + e ^ { x } } d x \)

To solve the integral \( \int \frac{1}{1 + e^x} \, dx \), we can make a clever substitution. If we set \( u = e^x \), then \( du = e^x \, dx = u \, dx \), which gives us \( dx = \frac{du}{u} \). The integral then becomes: \[ \int \frac{1}{1 + u} \frac{du}{u} = \int \frac{1}{u(1 + u)} \, du \] Now we can use partial fraction decomposition: \[ \frac{1}{u(1+u)} = \frac{A}{u} + \frac{B}{1+u} \] Solving for \( A \) and \( B \), we find \( A = 1 \) and \( B = -1 \), leading us to: \[ \int \left( \frac{1}{u} - \frac{1}{1+u} \right) du = \ln|u| - \ln|1+u| + C = \ln \left( \frac{u}{1+u} \right) + C \] Substituting back \( u = e^x \), our final result is: \[ \int \frac{1}{1 + e^x} \, dx = \ln \left( \frac{e^x}{1 + e^x} \right) + C = x - \ln(1 + e^x) + C \]