মান নির্ণয় করঃ  ∫₁^√3  dx/1+x²

Explanation:

Another Explanation (5): bài toán: \(\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}\) 🧐 আমরা জানি, \(\int \frac{1}{1+x^2} dx = \tan^{-1}(x) + C\) 😁 সুতরাং, \(\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2} = \left[ \tan^{-1}(x) \right]_{1}^{\sqrt{3}}\) 🤓 এখন, \(\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\) 🤩 এবং \(\tan^{-1}(1) = \frac{\pi}{4}\) 😎 অতএব, \(\left[ \tan^{-1}(x) \right]_{1}^{\sqrt{3}} = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) = \frac{\pi}{3} - \frac{\pi}{4}\) 🥳 \(= \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}\) 🎉 সুতরাং, \(\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2} = \frac{\pi}{12}\) 💖