\( \int\displaystyle\frac{1}{e^x-1}\,dx \) is equal to:
(where C is an arbitrary constant)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \log_e\left|\displaystyle\frac{e^x-1}{e^x}\right|+c \)
Dividing the numerator and the denominator by \( e^x \) makes the numerator the derivative of the denominator.
\( \displaystyle \int\frac{1}{e^x-1}\,dx=\int\frac{e^{-x}}{1-e^{-x}}\,dx \)
Let \( t=1-e^{-x} \), so \( dt=e^{-x}\,dx \):
\( \displaystyle \int\frac{dt}{t}=\log_e|t|+c \)
\( \displaystyle =\log_e\left|1-e^{-x}\right|+c \)
\( \displaystyle =\log_e\left|\frac{e^x-1}{e^x}\right|+c \)
Shortcut: Differentiate the options instead of integrating. Since \( \log_e\left|\frac{e^x-1}{e^x}\right|=\log_e|e^x-1|-x \),
\( \displaystyle \frac{d}{dx}\left[\log_e|e^x-1|-x\right]=\frac{e^x}{e^x-1}-1 \)
\( \displaystyle =\frac{1}{e^x-1} \)
Why the other options are incorrect
Option (1) \( -(e^x-1)^{-2}(e^x)+c \) — Incorrect. \( -(e^x-1)^{-2}(e^x) \) is the derivative of \( \frac{1}{e^x-1} \), not its integral.
Option (2) \( \log_e\left|\displaystyle\frac{e^x}{e^x-1}\right|+c \) — Incorrect. This is the negative of the correct answer, since \( \log_e\left|\frac{e^x}{e^x-1}\right|=-\log_e\left|\frac{e^x-1}{e^x}\right| \); its derivative is \( -\frac{1}{e^x-1} \).
Option (4) \( e^x+c \) — Incorrect. The derivative of \( e^x \) is \( e^x \), not \( \frac{1}{e^x-1} \).