Explanation: 
Another Explanation (5):
সমাধান:
ধরি, \(x = \sec{\theta}\) 🤓
তাহলে, \(dx = \sec{\theta} \tan{\theta} d\theta\) 😎
এখন, \(\sqrt{x^2 - 1} = \sqrt{\sec^2{\theta} - 1} = \sqrt{\tan^2{\theta}} = \tan{\theta}\) 🤩
অতএব,
\(\int \frac{dx}{x\sqrt{x^2-1}} = \int \frac{\sec{\theta} \tan{\theta} d\theta}{\sec{\theta} \tan{\theta}}\) 🥳
\(= \int d\theta\) 🤯
\(= \theta + C\) 😴
যেহেতু \(x = \sec{\theta}\), তাই \(\theta = \sec^{-1}{x}\) 🥰
সুতরাং, \(\int \frac{dx}{x\sqrt{x^2-1}} = \sec^{-1}{x} + C\) 🤗
