## Integrating With Factors

How do you evaluate
$\int \frac{1}{1+e^x}dx$
?
Sometimes it is possible to multiple by a factor equal to 1 so that the integration becomes simpler. In this case we can multiply the integrand by
$\frac{e^{-x}}{e^{-x}}$
to obtain
$\int \frac{e^{-x}}{e^{-x}+1} dx$
.
The numerator is almost the derivative of the denominator. We can write the integral; as
$- \int \frac{-e^{-x}}{e^{-x}+1} dx$
, then the numerator is the derivative of the denominator, and we can integrate by substituting
$u={e^{-x}+1}$
then
$du=-e^{-x}dx \rightarrow dx=-e^xdu$
.
.
\begin{aligned} - \int \frac{-e^{-x}}{e^{-x}+1} dx &=- \int \frac{e^{-x}}{u} e^x dx \\ &= - \int \frac{1}{u}du \\ &= -ln u +c \\ &= -ln(1+e^{-x})+c \end{aligned}

Integrating in this way can be tricky as the factor to be used is often not easy to find.