Answer:
[tex]\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx}= x - \ln(1 + e^{x}) + C\end{aligned}[/tex].
Step-by-step explanation:
The first derivative of the denominator [tex]1 + e^{x}[/tex] is [tex]e^{x}[/tex]. Rewrite the fraction to obtain that expression on the numerator.
[tex]\begin{aligned}\frac{1}{1 + e^{x}} &= \frac{1 + e^{x}}{1 + e^{x}} - \frac{e^{x}}{1+e^{x}}\\&=1-\frac{e^{x}}{1+e^{x}}\end{aligned}[/tex].
In other words,
[tex]\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\end{aligned}[/tex].
Apply [tex]u[/tex]-substitution on the integral [tex]\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx}[/tex]:
Let [tex]u = 1 + e^{x}[/tex]. [tex]u > 1[/tex].
[tex]du = e^{x}\cdot dx[/tex].
[tex]\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx} = \int{\frac{du}{u}} = \ln{|u|} = \ln{u} +C = \ln{(1+e^{x})}+C[/tex].
Therefore
[tex]\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\\ & = x - \ln{(1 + e^{x})}+C\end{aligned}[/tex].
