13 Multiple Choice 1 point Determine if the integral diverges or converges. Evaluate the integral if it converges. \( \int_{0}^{\infty} \frac{1}{e^{x}+e^{-x}} d x \)

To evaluate the integral \[ \int_{0}^{\infty} \frac{1}{e^{x}+e^{-x}} d x, \] we can simplify the integrand first. Notice that \[ e^{x} + e^{-x} = 2 \cosh(x), \] thus, we can rewrite the integral as: \[ \int_{0}^{\infty} \frac{1}{2 \cosh(x)} d x = \frac{1}{2} \int_{0}^{\infty} \frac{1}{\cosh(x)} d x. \] Next, we can evaluate \( \int_{0}^{\infty} \frac{1}{\cosh(x)} d x \). The integral \[ \int_{0}^{\infty} \frac{1}{\cosh(x)} d x \] is known to converge, and its value is \( \frac{\pi}{2} \). Therefore, we can conclude: \[ \int_{0}^{\infty} \frac{1}{\cosh(x)} d x = \frac{\pi}{2} \implies \frac{1}{2} \int_{0}^{\infty} \frac{1}{\cosh(x)} d x = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}. \] Thus, the original integral converges, and the value is \[ \int_{0}^{\infty} \frac{1}{e^{x}+e^{-x}} d x = \frac{\pi}{4}. \]