int_0^1(dx)/(e^x+e^-x) =?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int_{0}^{1} \frac{dx}{e^x + e^{-x}}\) \(I = \int_{0}^{1} \frac{e^x}{e^{2x} + 1} dx\) 😃 এখন, \(e^x = t\) ধরি। সুতরাং, \(e^x dx = dt\) যখন \(x = 0\), \(t = e^0 = 1\) এবং যখন \(x = 1\), \(t = e^1 = e\) তাহলে, \(I = \int_{1}^{e} \frac{dt}{t^2 + 1}\) 🤩 আমরা জানি, \(\int \frac{1}{x^2 + 1} dx = \tan^{-1}(x) + C\) সুতরাং, \(I = [\tan^{-1}(t)]_{1}^{e}\) 🤓 \(I = \tan^{-1}(e) - \tan^{-1}(1)\) 🥰 আমরা জানি, \(\tan^{-1}(1) = \frac{\pi}{4}\) সুতরাং, \(I = \tan^{-1}(e) - \frac{\pi}{4}\) 😎 অতএব, \(\int_{0}^{1} \frac{dx}{e^x + e^{-x}} = \tan^{-1}(e) - \frac{\pi}{4}\) 🥳