Question

$\int \frac{dx}{1+e^{-x}}$ is equal to
where c is the constant of integration

• A. $1+e^x+c$
• B. $\ln (1+e^{-x})+c$
• C. $\ln (1+e^x)+c$
• D. $2\ln (1+e^{-x})+c$

Answer 1

The correct option is C $\ln (1+e^x)+c$

Solution:

Let $I=\int \frac{dx}{1+e^{-x}}=\int \frac{dx}{1+\frac{1}{e^x}}=\int \frac{e^{x}dx}{1+e^x}$

Put $t=1+e^x$

$⟹dt=e^{x}dx$

Now, $I=\int \frac{dt}{t}=\ln t+C=\ln (1+e^x)+C$

Hence, C is the correct option.