How do you solve #\int \frac { e ^ { x } - 1} { e ^ { x } + 1} d x#?

2 Answers
Feb 10, 2017

The integral equals #x - 2ln|e^x/(e^x + 1)| + C#

Explanation:

By partial fractions:

#A/1 + B/(e^x + 1) = e^x - 2 + 1#

#A(e^x + 1) + B = e^x- 1#

#Ae^x + A + B = e^x- 1#

#Ae^x + (A + B) = e^x - 1#

We can write a system of equations.

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

Solving, we get #A = 1# and #B = -2#. The integral becomes:

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

We can separate.

#=>int 1dx - 2int1/(e^x + 1)dx#

Let #u = e^x#. Then #du = e^xdx#. Then #dx = (du)/e^x = (du)/u#

#=>int 1dx - 2int 1/((u + 1)u)du#

We're going to use partial fractions again.

#A/(u + 1) + B/u = 1#

#Au + Bu + B = 1#

Write a system of equations

#{(A + B = 0), (B = 1):}#

Solve to get #A = -1# and #B = 1#.

The expression becomes:

#=>int1dx - 2(int 1/udu - int1/( u + 1)du)#

#=>int1dx - 2int1/udu + 2int 1/(u + 1)du#

This can be readily integrated.

#=>x - 2ln|u| + 2ln|u +1| + C#

#=>x - 2(ln|u| - ln|u + 1|) + C#

The logarithms can be combined.

#=>x - 2ln|u/(u + 1)| + C#

#=>x - 2ln|e^x/(e^x + 1)| + C#

Hopefully this helps!

Apr 16, 2017

#-x+2ln(e^x+1)+C#

Explanation:

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

#color(white)(int(e^x-1)/(e^x+1)dx)=int(e^x+1)/(e^x+1)dx-2int1/(e^x+1)dx#

#color(white)(int(e^x-1)/(e^x+1)dx)=intdx-2int1/(e^x+1)dx#

#color(white)(int(e^x-1)/(e^x+1)dx)=x-2int1/(e^x+1)dx#

#color(white)(int(e^x-1)/(e^x+1)dx)=x-2int(e^x+1-e^x)/(e^x+1)dx#

#color(white)(int(e^x-1)/(e^x+1)dx)=x-2int(e^x+1)/(e^x+1)dx+2inte^x/(e^x+1)dx#

#color(white)(int(e^x-1)/(e^x+1)dx)=x-2intdx+2inte^x/(e^x+1)dx#

#color(white)(int(e^x-1)/(e^x+1)dx)=x-2x+2inte^x/(e^x+1)dx#

#color(white)(int(e^x-1)/(e^x+1)dx)=-x+2inte^x/(e^x+1)dx#

With the substitution #u=e^x+1# so #du=e^xdx#:

#color(white)(int(e^x-1)/(e^x+1)dx)=-x+2int1/udu#

#color(white)(int(e^x-1)/(e^x+1)dx)=-x+2lnabsu#

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

This is equivalent to the other answer provided by HSBC244! Note that:

#x-2lnabs(e^x/(e^x+1))=x-2(lnabs(e^x)-lnabs(e^x+1))#

#color(white)(x-2lnabs(e^x/(e^x+1)))=x-2x+2lnabs(e^x+1)#

#color(white)(x-2lnabs(e^x/(e^x+1)))=-x+2ln(e^x+1)#