int frac{d x}{left(e^x+e^{-x}right)^2}= | vedclass

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(B) Let $I = \int \frac{d x}{\left(e^x+e^{-x}\right)^2} = \int \frac{d x}{\left(e^x+\frac{1}{e^x}\right)^2}$ $= \int \frac{e^{2 x} d x}{\left(e^{2 x}+

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \frac{d x}{\left(e^x+e^{-x}\right)^2} = \int \frac{d x}{\left(e^x+\frac{1}{e^x}\right)^2}$ $= \int \frac{e^{2 x} d x}{\left(e^{2 x}+1\right)^2}$ Let $t = e^{2 x}+1$,then $dt = 2e^{2 x} dx$,which implies $e^{2 x} dx = \frac{dt}{2}$. Substituting these into the integral: $I = \int \frac{1}{2} \frac{dt}{t^2} = \frac{1}{2} \int t^{-2} dt$ $= \frac{1}{2} \left( \frac{t^{-1}}{-1} \right) + c = -\frac{1}{2t} + c$ Substituting $t = e^{2 x}+1$ back: $I = -\frac{1}{2\left(e^{2 x}+1\right)} + c$

$\int \frac{d x}{\left(e^x+e^{-x}\right)^2}=$

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