Integrating With Factors
\[\int \frac{1}{1+e^x}dx\]
?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 \[du=-e^{-x}dx \rightarrow dx=-e^xdu\]
..
\[\begin{equation} \begin{aligned} - \int \frac{-e^{-x}}{e^{-x}+1} dx &=- \int \frac{e^{-x}}{u} e^x dx \\ &= - \int \frac{1}{u}du \\ &= -ln u +c \\ &= -ln(1+e^{-x})+c \end{aligned} \end{equation} \]
Integrating in this way can be tricky as the factor to be used is often not easy to find.
