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 $