KCET · Maths · Indefinite Integration
\( \int \frac{1}{1+e^{x}} d x \) is equal to
- A \( \log _{e}\left(\frac{e^{x}+1}{e^{x}}\right)+0 \)
- B \( \log _{e}\left(\frac{e^{x}-1}{e^{x}}\right)+0 \)
- C \( \log _{e}\left(\frac{e^{x}}{e^{x}+1}\right)+0 \)
- D \( \log _{e}\left(\frac{e^{x}}{e^{x}-1}\right)+0 \)
Answer & Solution
Correct Answer
(C) \( \log _{e}\left(\frac{e^{x}}{e^{x}+1}\right)+0 \)
Step-by-step Solution
Detailed explanation
Given that
\[
\begin{aligned}
\int & \frac{1}{1+e^{x}} d x \\
=&-\int \frac{-e^{-x}}{e^{-x}+1} d x \\
=&-\ln \left|1+e^{-x}\right|+C \\
=&-\ln \left|\frac{e^{x}+1}{e^{x}}\right|+C \\
=& \ln \left|\frac{e^{x}}{e^{x}+1}\right|+C
\end{aligned}
\]
\[
\begin{aligned}
\int & \frac{1}{1+e^{x}} d x \\
=&-\int \frac{-e^{-x}}{e^{-x}+1} d x \\
=&-\ln \left|1+e^{-x}\right|+C \\
=&-\ln \left|\frac{e^{x}+1}{e^{x}}\right|+C \\
=& \ln \left|\frac{e^{x}}{e^{x}+1}\right|+C
\end{aligned}
\]
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