int frac{1}{(e^x + e^{-x})^2} , dx = | vedclass

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Question

(A) We have the integral $I = \int \frac{1}{(e^x + e^{-x})^2} \, dx$. Multiply the numerator and denominator by $e^{2x}$: $I = \int \frac{e^{2x}}{(e^x

Vedclass · Accepted Answer

$-\frac{1}{2(e^{2x} + 1)} + c$

Vedclass · Answer

(A) We have the integral $I = \int \frac{1}{(e^x + e^{-x})^2} \, dx$. Multiply the numerator and denominator by $e^{2x}$: $I = \int \frac{e^{2x}}{(e^x \cdot e^x + e^x \cdot e^{-x})^2} \, dx = \int \frac{e^{2x}}{(e^{2x} + 1)^2} \, dx$. Let $t = e^{2x} + 1$. Then $dt = 2e^{2x} \, dx$,which implies $e^{2x} \, dx = \frac{1}{2} \, dt$. Substituting these into the integral: $I = \int \frac{1}{t^2} \cdot \frac{1}{2} \, dt = \frac{1}{2} \int t^{-2} \, dt$. Integrating,we get $I = \frac{1}{2} \left( \frac{t^{-1}}{-1} \right) + c = -\frac{1}{2t} + c$. Substituting back $t = e^{2x} + 1$,we get $I = -\frac{1}{2(e^{2x} + 1)} + c$.

$\int \frac{1}{(e^x + e^{-x})^2} \, dx = $

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