Question

$\int \frac{1}{1+e^x}dx$ is equal to

• A. ${\log }_e\left(\frac{e^x+1}{e^x}\right)+c$
• B. ${\log }_e\left(\frac{e^x-1}{e^x}\right)+c$
• C. ${\log }_e\left(\frac{e^x}{e^x+1}\right)+c$
• D. ${\log }_e\left(\frac{e^x}{e^x-1}\right)+c$

Answer 1

Answer: (C)

$\int \frac{dx}{1+e^x}=\int \frac{dx}{1+\frac{1}{e^{-x}}}=-\int -\frac{e^{-x}}{e^{-x}+1}dx$
$=-\log \left(e^{-x}+1\right)+c=-\log \left(\frac{1}{e^x}+1\right)+c=-\log \left(\frac{1+e^x}{e^x}\right)+c$
$=\log \left(\frac{e^x}{1+e^x}\right)+c$