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(B) Let $I = \int \frac{d x}{e^x-1}$.Multiply the numerator and denominator by $e^{-x}$:$I = \int \frac{e^{-x}}{1-e^{-x}} d x$.Let $u = 1-e^{-x}$. The

Vedclass · Accepted Answer

$\log \left(e^x-1\right)-x+c, \quad$ where $c$ is the constant of integration.

Vedclass · Answer

(B) Let $I = \int \frac{d x}{e^x-1}$.Multiply the numerator and denominator by $e^{-x}$:$I = \int \frac{e^{-x}}{1-e^{-x}} d x$.Let $u = 1-e^{-x}$. Then $du = e^{-x} d x$.Substituting these into the integral:$I = \int \frac{1}{u} d u = \log |u| + c = \log |1-e^{-x}| + c$.Since $1-e^{-x} = \frac{e^x-1}{e^x}$,we have:$I = \log \left| \frac{e^x-1}{e^x} \right| + c = \log |e^x-1| - \log |e^x| + c$.$I = \log |e^x-1| - x + c$.

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

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