Q117 / 180 · KCET 2018 · Mathematics
\( \int \frac{1}{1+e^{x}} d x \) is equal to
- A\( \log _{e}\left(\frac{e^{x}+1}{e^{x}}\right)+0 \)
- B\( \log _{e}\left(\frac{e^{x}-1}{e^{x}}\right)+0 \)
- Correct answerC\( \log _{e}\left(\frac{e^{x}}{e^{x}+1}\right)+0 \)
- D\( \log _{e}\left(\frac{e^{x}}{e^{x}-1}\right)+0 \)
Solution
Given that
\[ \begin{aligned} \int & \frac{1}{1+e^{x}} d x \\ =&-\int \frac{-e^{-x}}{e^{-x}+1} d x \\ =&-\ln \left|1+e^{-x}\right|+C \\ =&-\ln \left|\frac{e^{x}+1}{e^{x}}\right|+C \\ =& \ln \left|\frac{e^{x}}{e^{x}+1}\right|+C \end{aligned} \]
\[ \begin{aligned} \int & \frac{1}{1+e^{x}} d x \\ =&-\int \frac{-e^{-x}}{e^{-x}+1} d x \\ =&-\ln \left|1+e^{-x}\right|+C \\ =&-\ln \left|\frac{e^{x}+1}{e^{x}}\right|+C \\ =& \ln \left|\frac{e^{x}}{e^{x}+1}\right|+C \end{aligned} \]
Hence, the correct answer is option: 'C'.