## 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 \[du=-e^{-x}dx \rightarrow dx=-e^xdu\]

..

\[\begin{equation} \begin{aligned} - \int \frac{-e^{-x}}{e^{-x}+1} dx &=- \int \frac{e^{-x}}{u} e^x dx \\ &= - \int \frac{1}{u}du \\ &= -ln u +c \\ &= -ln(1+e^{-x})+c \end{aligned} \end{equation} \]

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