Write a value

given $\int \frac{1}{1+e^{x}} d x$

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

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

Differentiating on both sides we get,

$E^{x} d x=d t$

Substituting above equation in given equation we get,

$=\int\left(1-\frac{1}{t}\right) d t$

$=t-\log t+c$

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