Question:
The value of integral \(\int _0^{\infty}\frac{1}{1+e^x}\,dx\) is...
The value of integral \(\int _0^{\infty}\frac{1}{1+e^x}\,dx\) is...
Show Hint
Multiply top and bottom by e^(-x) to get an integrand of the form -f'/f.
Updated On: Oct 1, 2026
- \(log2\)
- \(log(\frac{2}{e})\)
- \(-loge\)
- \(log(\frac{4}{e})\)
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The Correct Option is A
Solution and Explanation
Step 1: Understand the concept
The upper limit is infinity, so this is an improper integral. We first find an antiderivative and then take the limit.
Step 2: Rewrite the integrand
Multiply the numerator and denominator by \(e^{-x}\):
\[ \frac{1}{1 + e^x} = \frac{e^{-x}}{e^{-x} + 1} \]
The numerator is the negative of the derivative of the denominator.
Step 3: Antiderivative
\[ \int\frac{e^{-x}}{1 + e^{-x}}\,dx = -\ln(1 + e^{-x}) \]
Step 4: Apply the limits
At \(\infty\), \(e^{-x} \to 0\) so the value is \(-\ln1 = 0\). At \(0\), the value is \(-\ln2\). So the integral is \(0 - (-\ln 2) = \ln 2 = \log 2\), option (A).
Final Answer:
The integral converges to log 2. This is option (A). \[ \boxed{\text{(A) }\log 2} \]
The upper limit is infinity, so this is an improper integral. We first find an antiderivative and then take the limit.
Step 2: Rewrite the integrand
Multiply the numerator and denominator by \(e^{-x}\):
\[ \frac{1}{1 + e^x} = \frac{e^{-x}}{e^{-x} + 1} \]
The numerator is the negative of the derivative of the denominator.
Step 3: Antiderivative
\[ \int\frac{e^{-x}}{1 + e^{-x}}\,dx = -\ln(1 + e^{-x}) \]
Step 4: Apply the limits
At \(\infty\), \(e^{-x} \to 0\) so the value is \(-\ln1 = 0\). At \(0\), the value is \(-\ln2\). So the integral is \(0 - (-\ln 2) = \ln 2 = \log 2\), option (A).
Final Answer:
The integral converges to log 2. This is option (A). \[ \boxed{\text{(A) }\log 2} \]
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