Question #68d98

3 Answers
Sep 12, 2017

The answer is 2ln(e^{x}+1)-x+C

Explanation:

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)

Sep 12, 2017

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

Explanation:

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

Sep 12, 2017

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

Explanation:

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.