Question

Question #68d98

Answer 1

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)`

Answer 2

`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`

Answer 3

`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.