How do you integrate #(1 + e^x )/(1 - e^x)#?

Question

35888 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-19T23:41:42

#int (1+e^x)/(1-e^x)dx = x - 2ln abs(1-e^x)+C#

Substitute:

#t = e^x #

#dt = e^xdx#

#dx = dt/t#

so:

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

Use partial fractions decomposition:

#(1+t)/(t(1-t)) = A/t +B/(1-t)#

#(1+t)/(t(1-t)) = (A(1-t)+Bt)/(t(1-t))#

#1+t= A-At+Bt#

#{(A=1),(-A+B=1):}#

#{(A=1),(B=2):}#

so:

#int (1+e^x)/(1-e^x)dx = int dt/t +2 int dt/(1-t)#

#int (1+e^x)/(1-e^x)dx = ln abs t - 2ln abs(1-t)+C#

and undoing the substitution:

#int (1+e^x)/(1-e^x)dx = x - 2ln abs(1-e^x)+C#