How do you integrate # 1/(1+e^x) # using partial fractions?

Question

William Abdalla · Accepted Answer

#int1/(1+e^x)dx = x-ln(1+e^x)+C#

We use substitution to put the problem into a form where we can use partial fractions.

Let #u = 1+e^x => du = e^xdx#. Then:

#int1/(1+e^x)dx = inte^x/(e^x(1+e^x)dx#

#=int1/(u(u-1)du#

Decomposing #1/(u(u-1))#, we find

#1/(u(u-1)) = 1/(u-1) - 1/u#

#=> int1/(u(u-1))du = int1/(u-1)du - int1/udu#

#=ln|u-1|-ln|u|+C#

#=ln|e^x+1-1|-ln|e^x+1|+C#

#=ln(e^x)-ln(e^x+1)+C#

#=x-ln(e^x+1)+C#

Rather than using partial fractions, we can also solve this with a little algebraic manipulation and substitution.

#int1/(1+e^x)dx = int(1+e^x-e^x)/(1+e^x)dx#

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

#=intdx-inte^x/(1+e^x)dx#

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

Focusing on the remaining integral, let #u = (1+e^x) => du = e^xdx#

Substituting, we have

#inte^x/(1+e^x)dx = int1/udu#

#=ln|u|+C#

#=ln(1+e^x)+C#

Plugging this into the equation above, we can get our result:

#int1/(1+e^x)dx = x-ln(1+e^x)+C#

Elijah Abram · Answer

undefined To integrate $ \frac{1}{1+e^x} $ using partial fractions, we first express the fraction as a sum of simpler fractions. Since the denominator is a sum of two terms, $1$ and $e^x$, we write $ \frac{1}{1+e^x} $ as $ \frac{A}{1} + \frac{B}{e^x} $. Multiplying both sides by $1+e^x$ to clear the denominators, we get $ 1 = A(e^x) + B(1) $.

To solve for $ A $ and $ B $, we let $ x = 0 $ to eliminate the term involving $ e^x $, yielding $ 1 = A + B $. Then, we let $ x = -\infty $ to eliminate the term involving $ 1 $, which gives us $ 1 = 0 + B $. Solving these equations simultaneously, we find that $ B = 1 $ and $ A = 0 $.

Thus, we have $ \frac{1}{1+e^x} = \frac{1}{e^x} $. Integrating both sides, we get $ \int \frac{1}{1+e^x} \, dx = \int \frac{1}{e^x} \, dx $.

The integral of $ \frac{1}{e^x} $ with respect to $ x $ is simply $ -e^{-x} + C $, where $ C $ is the constant of integration.

Therefore, $ \int \frac{1}{1+e^x} \, dx = -e^{-x} + C $.