int frac{1}{sqrt{1 - e^{2x}}} , dx = | vedclass

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Question

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

Vedclass · Accepted Answer

$x - \log[1 + \sqrt{1 - e^{2x}}] + c$

Vedclass · Answer

(A) Let $I = \int \frac{1}{\sqrt{1 - e^{2x}}} \, dx$. Multiply the numerator and denominator by $e^{-x}$: $I = \int \frac{e^{-x}}{\sqrt{e^{-2x} - 1}} \, dx$. Substitute $t = e^{-x}$,then $dt = -e^{-x} \, dx$,which implies $e^{-x} \, dx = -dt$. $I = -\int \frac{1}{\sqrt{t^2 - 1}} \, dt$. Using the standard integral formula $\int \frac{1}{\sqrt{t^2 - a^2}} \, dt = \log|t + \sqrt{t^2 - a^2}| + c$: $I = -\log|t + \sqrt{t^2 - 1}| + c$. Substitute $t = e^{-x}$ back: $I = -\log|e^{-x} + \sqrt{e^{-2x} - 1}| + c$. $I = -\log|\frac{1}{e^x} + \frac{\sqrt{1 - e^{2x}}}{e^x}| + c$. $I = -\log|\frac{1 + \sqrt{1 - e^{2x}}}{e^x}| + c$. $I = -\log(1 + \sqrt{1 - e^{2x}}) + \log(e^x) + c$. $I = x - \log(1 + \sqrt{1 - e^{2x}}) + c$.

$\int \frac{1}{\sqrt{1 - e^{2x}}} \, dx = $

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