Evaluate the following integral or state that it diverges. \( \int_{0}^{\infty} \frac{16 \tan ^{-1} v}{1+v^{2}} d v \) Select the correct choice and, if necessary, fill in the answer box to complete your choice. A. The improper integral converges and \( \int_{0}^{\infty} \frac{16 \tan ^{-1} v}{1+v^{2}} d v=\square \). (Type an exact answer in terms of \( \pi \).)

Let \[ I = \int_{0}^{\infty} \frac{16 \tan^{-1} v}{1+v^2} \, dv. \] **Step 1. Factor out the constant:** \[ I = 16 \int_{0}^{\infty} \frac{\tan^{-1} v}{1+v^2} \, dv. \] **Step 2. Use the substitution:** Let \[ u = \tan^{-1} v. \] Then, \[ v = \tan u \quad \text{and} \quad dv = \sec^2 u \, du. \] Notice that \[ 1 + v^2 = 1 + \tan^2 u = \sec^2 u. \] **Step 3. Change the limits of integration:** When \(v = 0\), \[ u = \tan^{-1}(0) = 0. \] When \(v \to \infty\), \[ u = \tan^{-1}(v) \to \frac{\pi}{2}. \] **Step 4. Substitute into the integral:** \[ \int_{0}^{\infty} \frac{\tan^{-1} v}{1+v^2} \, dv = \int_{0}^{\pi/2} \frac{u}{\sec^2 u} \cdot \sec^2 u\,du = \int_{0}^{\pi/2} u \, du. \] **Step 5. Evaluate the integral:** \[ \int_{0}^{\pi/2} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{\pi/2} = \frac{(\pi/2)^2}{2} = \frac{\pi^2}{8}. \] **Step 6. Multiply by the constant:** \[ I = 16 \cdot \frac{\pi^2}{8} = 2\pi^2. \] Thus, the integral converges and \[ \int_{0}^{\infty} \frac{16 \tan^{-1} v}{1+v^2} \, dv = 2\pi^2. \]