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

Question

Emily Ammerman · Accepted Answer

#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#

Charles Anctil · Answer

undefined To integrate $ \frac{1 + e^x}{1 - e^x} $, you can use the substitution method. Let $ u = e^x $, then $ du = e^x dx $. Substitute these into the integral:

$$ \int \frac{1 + e^x}{1 - e^x} dx = \int \frac{1 + u}{1 - u} du $$

Now, perform the partial fraction decomposition on the integrand:

$$ \frac{1 + u}{1 - u} = 1 + \frac{2u}{1 - u} $$

$$ = 1 + \frac{2u}{(1 - u)(1 + u)} $$

$$ = 1 + \frac{2u}{1 - u^2} $$

Now, rewrite the integral:

$$ \int \left(1 + \frac{2u}{1 - u^2}\right) du $$

$$ = \int du + \int \frac{2u}{1 - u^2} du $$

The first integral is straightforward:

$$ = u + \int \frac{2u}{1 - u^2} du $$

For the second integral, use a substitution $ w = u^2 $:

$$ \int \frac{2u}{1 - u^2} du = \int \frac{2}{1 - w} dw $$

$$ = -2 \ln|1 - w| + C $$

Substitute back $ w = u^2 $:

$$ = -2 \ln|1 - u^2| + C $$

Finally, integrate $ u $ back into the equation:

$$ = u - 2 \ln|1 - u^2| + C $$

Substitute back $ u = e^x $:

$$ = e^x - 2 \ln|1 - e^{2x}| + C $$

So, the integral of $ \frac{1 + e^x}{1 - e^x} $ is $ e^x - 2 \ln|1 - e^{2x}| + C $, where $ C $ is the constant of integration.