We are tasked with solving the integral: \[ \int \frac{dx}{1 + e^x} \] This integral is of a standard form. We recognize that the integral of \( \frac{1}{1 + e^x} \) is directly related to the natural logarithm function.
Specifically: \[ \int \frac{dx}{1 + e^x} = \log|1 + e^x| + C \] where \( C \) is the constant of integration. Thus, the correct answer is \( \log|1 + e^x| + C \).
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]