intdx/(e^x+e^-x)^2=?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int \frac{dx}{(e^x + e^{-x})^2}\) \(I = \int \frac{dx}{(e^x + \frac{1}{e^x})^2}\) \(I = \int \frac{dx}{(\frac{e^{2x} + 1}{e^x})^2}\) \(I = \int \frac{e^{2x} dx}{(e^{2x} + 1)^2}\) এখন, ধরি \(u = e^{2x} + 1\) তাহলে, \(\frac{du}{dx} = 2e^{2x}\) সুতরাং, \(dx = \frac{du}{2e^{2x}}\) তাহলে, \(I = \int \frac{e^{2x}}{u^2} \cdot \frac{du}{2e^{2x}}\) \(I = \int \frac{1}{2u^2} du\) \(I = \frac{1}{2} \int u^{-2} du\) \(I = \frac{1}{2} \cdot \frac{u^{-1}}{-1} + c\) \(I = -\frac{1}{2u} + c\) \(I = -\frac{1}{2(e^{2x} + 1)} + c\) সুতরাং, \(\int \frac{dx}{(e^x + e^{-x})^2} = -\frac{1}{2(e^{2x} + 1)} + c\) 🎉