Question #68d98

Start by letting #u=e^x+1# so that #du=e^{x}dx# and #e^x-1=u-2#. Then #dx=1/e^(x) du=1/(u-2)du# and the integral becomes

#\int (e^x-1)/(e^x+1)dx=\int (u-2)/(u(u-1))du#.

Next, use the Method of Partial Fractions to write #(u-2)/(u(u-1))# as #2/u-1/(u-1)#, which can be easily integrated to get #2ln|u|-ln|u-1|+C#.

Now substitute #u=e^x+1#, which is never negative, and use the fact that #ln(e^x)=x# to get

#\int (e^x-1)/(e^x+1)dx=2ln(e^{x}+1)-x+C#.

This can be checked by differentiation:

#d/dx(2ln(e^x+1)-x+C)=2/(e^x+1)*e^x-1=(2e^x-(e^x+1))/(e^x+1)=(e^x-1)/(e^x+1)#

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

Insert zero into the numerator:

#int(e^x+ 0-1)/(e^x+1)dx#

In place of the 0 we write #e^x-e^x#:

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

Now we do some clever grouping:

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

We combine the first group:

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

Separate into two fractions:

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

Please notice that the second fraction becomes 1:

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

Separate into two integrals:

#2inte^x/(e^x+1)dx -intdx#

For the first integral we let #u = e^x+1#, then #du = e^xdx#:

#2int1/udu -intdx#

We know these integrals very well:

#2ln|u| -x+ C#

Reverse the substitution:

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

We can also do this integral as follows:

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

Splitting up the fraction as #(e^x+1)/(e^x+1)-2/(e^x+1)#, this becomes:

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

We can integrate the first term easily:

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

Now, multiply the integrand by #e^-x/e^-x#. This seems ridiculous, but you'll see why it works in a second:

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

Let #u=1+e^-x#, implying that #du=-e^-xdx#. Then:

#=x+2int1/udu#

#=x+2lnabsu+C#

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

As #e^-x>0# for all Real values of #x#, the absolute value bars aren't needed.

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

Which can be shown to be equivalent to the other provided answers.