Evaluate the integral. \[ \int \frac{d x}{x^{2} \sqrt{x^{2}-25}}, x>5 \]

To evaluate the integral \[ \int \frac{d x}{x^{2} \sqrt{x^{2}-25}}, \quad x>5, \] we can make the substitution \( x = 5 \sec(\theta) \). Then, we have: \[ dx = 5 \sec(\theta) \tan(\theta) d\theta \] Next, we substitute \( x \) into the integral: \[ \sqrt{x^2 - 25} = \sqrt{(5 \sec(\theta))^2 - 25} = \sqrt{25 \sec^2(\theta) - 25} = \sqrt{25(\sec^2(\theta) - 1)} = \sqrt{25 \tan^2(\theta)} = 5 \tan(\theta). \] Now we have \( x^2 = (5 \sec(\theta))^2 = 25 \sec^2(\theta) \) and we can rewrite the integral: \[ \int \frac{dx}{x^2 \sqrt{x^2 - 25}} = \int \frac{5 \sec(\theta) \tan(\theta) d\theta}{25 \sec^2(\theta) \cdot 5 \tan(\theta)} = \int \frac{5 \sec(\theta) \tan(\theta)}{125 \sec^2(\theta) \tan(\theta)} d\theta = \frac{1}{25} \int \frac{d\theta}{\sec(\theta)} = \frac{1}{25} \int \cos(\theta) d\theta. \] Now, the integral of \( \cos(\theta) \) is: \[ \int \cos(\theta) d\theta = \sin(\theta) + C. \] Thus, substituting this back, we have: \[ \frac{1}{25}(\sin(\theta) + C) = \frac{1}{25} \sin(\theta) + \frac{C}{25}. \] Next, we need to express \( \sin(\theta) \) in terms of \( x \). From \( x = 5 \sec(\theta) \), we have: \[ \sec(\theta) = \frac{x}{5} \implies \cos(\theta) = \frac{5}{x}. \] We can find \( \sin(\theta) \) using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \): \[ \sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \left(\frac{5}{x}\right)^2 = 1 - \frac{25}{x^2} = \frac{x^2 - 25}{x^2}. \] Thus, \[ \sin(\theta) = \frac{\sqrt{x^2 - 25}}{x}. \] Putting it back into our integral gives us: \[ \frac{1}{25} \sin(\theta) = \frac{1}{25} \cdot \frac{\sqrt{x^2 - 25}}{x} = \frac{\sqrt{x^2 - 25}}{25x}. \] Finally, we add the constant back in, \[ \int \frac{dx}{x^{2} \sqrt{x^{2}-25}} = \frac{\sqrt{x^2 - 25}}{25x} + C. \] So the final answer is: \[ \boxed{\frac{\sqrt{x^2 - 25}}{25x} + C}. \]