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KCET · Maths · Indefinite Integration

\( \int \frac{1}{1+e^{x}} d x \) is equal to

  1. A \( \log _{e}\left(\frac{e^{x}+1}{e^{x}}\right)+0 \)
  2. B \( \log _{e}\left(\frac{e^{x}-1}{e^{x}}\right)+0 \)
  3. C \( \log _{e}\left(\frac{e^{x}}{e^{x}+1}\right)+0 \)
  4. D \( \log _{e}\left(\frac{e^{x}}{e^{x}-1}\right)+0 \)
Verified Solution

Answer & Solution

Correct Answer

(C) \( \log _{e}\left(\frac{e^{x}}{e^{x}+1}\right)+0 \)

Step-by-step Solution

Detailed explanation

Given that
\[
\begin{aligned}
\int & \frac{1}{1+e^{x}} d x \\
=&-\int \frac{-e^{-x}}{e^{-x}+1} d x \\
=&-\ln \left|1+e^{-x}\right|+C \\
=&-\ln \left|\frac{e^{x}+1}{e^{x}}\right|+C \\
=& \ln \left|\frac{e^{x}}{e^{x}+1}\right|+C
\end{aligned}
\]
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