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

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(A) Given integral: $I = \int \frac{dx}{e^{-2x}(e^{2x} + 1)^2}$ Since $\frac{1}{e^{-2x}} = e^{2x}$,the integral becomes: $I = \int \frac{e^{2x} dx}{(e

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

Vedclass · Answer

(A) Given integral: $I = \int \frac{dx}{e^{-2x}(e^{2x} + 1)^2}$ Since $\frac{1}{e^{-2x}} = e^{2x}$,the integral becomes: $I = \int \frac{e^{2x} dx}{(e^{2x} + 1)^2}$ Let $t = e^{2x} + 1$. Then,differentiating with respect to $x$,we get $dt = 2e^{2x} dx$,which implies $e^{2x} dx = \frac{dt}{2}$. Substituting these into the integral: $I = \int \frac{1}{t^2} \cdot \frac{dt}{2} = \frac{1}{2} \int t^{-2} dt$ Using the power rule for integration $\int t^n dt = \frac{t^{n+1}}{n+1}$: $I = \frac{1}{2} \left( \frac{t^{-1}}{-1} \right) + c = -\frac{1}{2t} + c$ Substituting back $t = e^{2x} + 1$: $I = \frac{-1}{2(e^{2x} + 1)} + c$

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

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