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

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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:

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

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