Evaluate the integral. \( \int_{2}^{5} \frac{d x}{\left(x^{2}-1\right)^{3 / 2}} \)

To evaluate the integral \[ I = \int_{2}^{5} \frac{dx}{(x^2 - 1)^{3/2}}, \] we can use a trigonometric substitution. Notice that the expression \((x^2 - 1)^{3/2}\) suggests using the substitution where \( x = \sec(\theta) \). With this substitution, we get: 1. \( dx = \sec(\theta) \tan(\theta) d\theta \) 2. \( x^2 - 1 = \sec^2(\theta) - 1 = \tan^2(\theta) \) Now the integral becomes: \[ I = \int \frac{\sec(\theta) \tan(\theta) \, d\theta}{(\tan^2(\theta))^{3/2}} = \int \frac{\sec(\theta) \tan(\theta) \, d\theta}{\tan^3(\theta)} = \int \frac{\sec(\theta)}{\tan^2(\theta)} \, d\theta. \] Since \(\sec(\theta) = \frac{1}{\cos(\theta)}\) and \(\tan^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)}\), we can rewrite the integral as: \[ I = \int \frac{\cos^2(\theta)}{\sin^2(\theta)} \, d\theta = \int \cot^2(\theta) \, d\theta. \] The integral of \(\cot^2(\theta)\) is given by: \[ \int \cot^2(\theta) \, d\theta = -\cot(\theta) + \theta + C. \] Now let's convert back to \(x\) from \(\theta\). Since \(x = \sec(\theta)\), we have: \[ \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sqrt{x^2 - 1}}. \] Also, \(\theta = \sec^{-1}(x)\). Thus, \[ I = -\frac{1}{\sqrt{x^2 - 1}} + \sec^{-1}(x). \] We'll evaluate this from \(x = 2\) to \(x = 5\): 1. At \(x = 5\): \[ -\frac{1}{\sqrt{5^2 - 1}} + \sec^{-1}(5) = -\frac{1}{\sqrt{24}} + \sec^{-1}(5), \] 2. At \(x = 2\): \[ -\frac{1}{\sqrt{2^2 - 1}} + \sec^{-1}(2) = -\frac{1}{\sqrt{3}} + \sec^{-1}(2). \] Thus, \[ I = \left(-\frac{1}{\sqrt{24}} + \sec^{-1}(5)\right) - \left(-\frac{1}{\sqrt{3}} + \sec^{-1}(2)\right). \] Calculating this gives: \[ I = \sec^{-1}(5) - \sec^{-1}(2) + \left(\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{24}}\right). \] Calculating \( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{24}} \): \[ = \frac{2}{2\sqrt{3}} - \frac{1}{\sqrt{24}} = \frac{2}{2\sqrt{3}} - \frac{1}{2\sqrt{6}}. \] To finalize: \[ I = \sec^{-1}(5) - \sec^{-1}(2) + \left(\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{24}}\right). \] This is the evaluated integral. Further simplification can be made by substituting the values of \( \sec^{-1}(5) \) and \( \sec^{-1}(2) \).