Question:
\[ \int \frac{1}{\mathrm{e}^x+1} \mathrm{~d} x= \]
\[ \int \frac{1}{\mathrm{e}^x+1} \mathrm{~d} x= \]
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Multiply the numerator and denominator by e^(-x) and substitute.
Updated On: Nov 13, 2026
- \(x+\mathrm{log}(\mathrm{e}^{x}+1)+\mathrm{c}\), where c is the constant of integration.
- \(x-\mathrm{log}(\mathrm{e}^{x}+1)+\mathrm{c}\), where c is the constant of integration.
- \(\mathrm{log}(\mathrm{e}^{x}-1)+x+\mathrm{c}\), where c is the constant of integration.
- \(\mathrm{log}(\mathrm{e}^{\mathrm{x}}-1)-x+\mathrm{c}\), where c is the constant of integration.
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Write \(\dfrac{1}{e^x+1} = \dfrac{e^{-x}}{1+e^{-x}}\) by multiplying the numerator and denominator by \(e^{-x}\).
Step 2: Substitute
Let \(t = 1 + e^{-x}\), so \(dt = -e^{-x}dx\).
Step 3: Integrate
\(\int \dfrac{e^{-x}}{1+e^{-x}}dx = -\int\dfrac{dt}{t} = -\log(1+e^{-x}) + c\).
Step 4: Simplify
\(-\log\dfrac{e^x+1}{e^x} = -\log(e^x+1) + x\).
So the integral is \(x - \log(e^x+1) + c\).
Final Answer:
The integral is x - log(e^x + 1) + c. \[ \boxed{\text{(B) }x-\log(\mathrm{e}^{x}+1)+\mathrm{c}} \]
Write \(\dfrac{1}{e^x+1} = \dfrac{e^{-x}}{1+e^{-x}}\) by multiplying the numerator and denominator by \(e^{-x}\).
Step 2: Substitute
Let \(t = 1 + e^{-x}\), so \(dt = -e^{-x}dx\).
Step 3: Integrate
\(\int \dfrac{e^{-x}}{1+e^{-x}}dx = -\int\dfrac{dt}{t} = -\log(1+e^{-x}) + c\).
Step 4: Simplify
\(-\log\dfrac{e^x+1}{e^x} = -\log(e^x+1) + x\).
So the integral is \(x - \log(e^x+1) + c\).
Final Answer:
The integral is x - log(e^x + 1) + c. \[ \boxed{\text{(B) }x-\log(\mathrm{e}^{x}+1)+\mathrm{c}} \]
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