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

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Question

(A) To solve the integral $I = \int \frac{dx}{e^x - 1}$,multiply the numerator and denominator by $e^{-x}$:$I = \int \frac{e^{-x}}{e^{-x}(e^x - 1)} dx

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To solve the integral $I = \int \frac{dx}{e^x - 1}$,multiply the numerator and denominator by $e^{-x}$:$I = \int \frac{e^{-x}}{e^{-x}(e^x - 1)} dx = \int \frac{e^{-x}}{1 - e^{-x}} dx$Let $t = 1 - e^{-x}$.Then,the derivative is $dt = e^{-x} dx$.Substituting these into the integral,we get:$I = \int \frac{dt}{t} = \ln|t| + c$Substituting back $t = 1 - e^{-x}$,we get:$I = \ln(1 - e^{-x}) + c$.

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

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