How do you use partial fraction decomposition to decompose the fraction to integrate #1/(1+e^x) #?

1 Answer
Oct 17, 2015

#int 1/(1+e^(x))\ dx=x-ln(1+e^(x))+C#.

Explanation:

This is not really a partial fraction problem, but instead it's just a trick. Write:

#1/(1+e^(x))=(1+e^(x)-e^(x))/(1+e^(x))#

#=(1+e^(x))/(1+e^(x))-e^(x)/(1+e^(x))=1-e^(x)/(1+e^(x))#

This means

#int 1/(1+e^(x))\ dx=int (1-e^(x)/(1+e^(x)))\ dx#

#=x-int e^(x)/(1+e^(x))\ dx#.

For this last integral, let #u=1+e^(x)# so that #du=e^(x)\ dx# and we get

#int e^(x)/(1+e^(x))\ dx=int\ (du)/u=ln|u|+C=ln(1+e^(x))+C# (note that #1+e^(x)>0# for all #x#).

Therefore,

#int 1/(1+e^(x))\ dx=x-ln(1+e^(x))+C#.