Question

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

• A. ${\log }_e\left(\frac{1+e^x}{e^x}\right)+c$
• B. ${\log }_e\left(\frac{e^x}{1+e^x}\right)+c$
• C. ${\log }_e\left(e^x\right)\left(e^x+1\right)+c$
• D. ${\log }_e\left(e^{2x}+1\right)+c$

Answer 1

The correct option is B ${\log }_e\left(\frac{e^x}{1+e^x}\right)+c$

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

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

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

Let $t=e^{-x}+1$

$=-\int \frac{dt}{t}$

$=-\log \left(t\right)+c$

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

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

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