Question:
\[ \int \frac{\mathrm{d} x}{\mathrm{e}^x-1}= \]
\[ \int \frac{\mathrm{d} x}{\mathrm{e}^x-1}= \]
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Multiply numerator and denominator by e to the power minus x, then integrate.
Updated On: Nov 13, 2026
- \(\mathrm{log}(\mathrm{e}^{\mathrm{x}}-1)+x+\mathrm{c}\), where c is the constant of integration.
- \(\mathrm{log}(\mathrm{e}^{\mathrm{x}}-1)-x+\mathrm{c}\), where c is the constant of integration.
- \(x-\mathrm{log}(\mathrm{e}^{x}-1)+\mathrm{c}\), where c is the constant of integration.
- \(\mathrm{log}(\mathrm{e}^{\mathrm{x}}-1)-x\mathrm{e}^{\mathrm{x}}+\mathrm{c}\), where c is the constant of integration.
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Rewrite \(\dfrac{1}{e^x - 1}\) as \(\dfrac{e^{-x}}{1 - e^{-x}}\) by dividing top and bottom by \(e^x\).
Step 2: Integrate
The numerator \(e^{-x}\) is the derivative of the denominator \(1 - e^{-x}\), so \(\displaystyle\int\dfrac{e^{-x}}{1 - e^{-x}}dx = \log(1 - e^{-x}) + c\).
Step 3: Rewrite the log
\(\log(1 - e^{-x}) = \log\dfrac{e^x - 1}{e^x} = \log(e^x - 1) - x\).
Final Answer:
The result is \(\log(e^x - 1) - x + c\). \[ \boxed{\text{(B) }\log(e^{x}-1) - x + c} \]
Rewrite \(\dfrac{1}{e^x - 1}\) as \(\dfrac{e^{-x}}{1 - e^{-x}}\) by dividing top and bottom by \(e^x\).
Step 2: Integrate
The numerator \(e^{-x}\) is the derivative of the denominator \(1 - e^{-x}\), so \(\displaystyle\int\dfrac{e^{-x}}{1 - e^{-x}}dx = \log(1 - e^{-x}) + c\).
Step 3: Rewrite the log
\(\log(1 - e^{-x}) = \log\dfrac{e^x - 1}{e^x} = \log(e^x - 1) - x\).
Final Answer:
The result is \(\log(e^x - 1) - x + c\). \[ \boxed{\text{(B) }\log(e^{x}-1) - x + c} \]
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