The value of the integral int frac{d x}{left( | vedclass

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(C) Let $I = \int \frac{dx}{(e^x + e^{-x})^2}$.Multiply the numerator and denominator by $e^{2x}$:$I = \int \frac{e^{2x} dx}{(e^x \cdot e^x + e^{-x} \

Vedclass · Accepted Answer

$-\frac{1}{2}\left(e^{2 x}+1\right)^{-1}+C$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{(e^x + e^{-x})^2}$.Multiply the numerator and denominator by $e^{2x}$:$I = \int \frac{e^{2x} dx}{(e^x \cdot e^x + e^{-x} \cdot e^x)^2} = \int \frac{e^{2x} dx}{(e^{2x} + 1)^2}$.Let $u = e^{2x} + 1$. Then $du = 2e^{2x} dx$, which implies $e^{2x} dx = \frac{du}{2}$.Substituting these into the integral:$I = \int \frac{du/2}{u^2} = \frac{1}{2} \int u^{-2} du$.Integrating with respect to $u$:$I = \frac{1}{2} \left( \frac{u^{-1}}{-1} \right) + C = -\frac{1}{2u} + C$.Substituting $u = e^{2x} + 1$ back:$I = -\frac{1}{2(e^{2x} + 1)} + C = -\frac{1}{2}(e^{2x} + 1)^{-1} + C$.

The value of the integral $\int \frac{d x}{\left(e^x+e^{-x}\right)^2}$ is

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