The value of \(\int_{0}^{1}\displaystyle \frac{1}{1 + e^{-x}}\,dx\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\log\displaystyle\frac{e + 1}{2}\)
Multiplying the numerator and the denominator by \(e^{x}\):
\( \displaystyle \frac{1}{1 + e^{-x}} = \frac{e^{x}}{e^{x} + 1} \)
Now the numerator is the derivative of the denominator, so \(\int \frac{e^{x}}{e^{x} + 1}\,dx = \log(e^{x} + 1)\):
\( \displaystyle \int_{0}^{1} \frac{e^{x}}{e^{x} + 1}\,dx = \Big[\log(e^{x} + 1)\Big]_{0}^{1} \)
\(= \log(e + 1) - \log 2\)
\( \displaystyle = \log\frac{e + 1}{2} \)
Why the other options are incorrect
Option (2) \(\log(e + 1)\) — Incorrect. Only the upper limit has been used; the value at x = 0 is \(\log(1 + 1) = \log 2\), not 0.
Option (3) \(\log 2\) — Incorrect. This is the value of \(\log(e^{x} + 1)\) at the lower limit only.
Option (4) \(\log\displaystyle\frac{2e}{e + 1}\) — Incorrect. This is \(\int_{0}^{1}\frac{1}{1 + e^{x}}\,dx = 1 - \log\frac{e + 1}{2}\); the sign of the power of e has been changed.