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The correct answer is Option (3) → \(\log\left(\displaystyle\frac{e^x - 1}{e^x}\right) + C\)

Dividing the numerator and the denominator by \(e^x\):

    \( \displaystyle \int\frac{dx}{e^x - 1} = \int\frac{e^{-x}}{1 - e^{-x}}\,dx \)

Now the numerator is the derivative of the denominator, since \(\frac{d}{dx}\left(1 - e^{-x}\right) = e^{-x}\). So

    \( \displaystyle \int\frac{e^{-x}}{1 - e^{-x}}\,dx = \log\left|1 - e^{-x}\right| + C \)

    \( \displaystyle = \log\left|\frac{e^x - 1}{e^x}\right| + C \)

Shortcut: Differentiate Option (3):

    \( \displaystyle \frac{d}{dx}\left[\log\left(e^x - 1\right) - x\right] = \frac{e^x}{e^x - 1} - 1 \)

    \( \displaystyle = \frac{1}{e^x - 1} \)