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

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

Vedclass · Accepted Answer

$\frac{1}{3}\log (e^x - 1) - \frac{1}{3}\log (e^x + 2) + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{e^x + 1 - 2e^{-x}}$.Multiply numerator and denominator by $e^x$:$I = \int \frac{e^x dx}{e^{2x} + e^x - 2}$.Let $e^x = t$,then $e^x dx = dt$.$I = \int \frac{dt}{t^2 + t - 2} = \int \frac{dt}{(t + 2)(t - 1)}$.Using partial fractions:$\frac{1}{(t + 2)(t - 1)} = \frac{1}{3} \left[ \frac{1}{t - 1} - \frac{1}{t + 2} \right]$.Integrating both sides:$I = \frac{1}{3} \int \left( \frac{1}{t - 1} - \frac{1}{t + 2} \right) dt = \frac{1}{3} \log |t - 1| - \frac{1}{3} \log |t + 2| + c$.Substituting $t = e^x$ back:$I = \frac{1}{3} \log (e^x - 1) - \frac{1}{3} \log (e^x + 2) + c$.

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

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