int dx/(xsqrt(x^2-1))=? 

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, \(\int \frac{dx}{x\sqrt{x^2-1}} = \sec^{-1}{x} + C\) 🤗