# Question #68d98

Sep 12, 2017

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

#### Explanation:

Start by letting $u = {e}^{x} + 1$ so that $\mathrm{du} = {e}^{x} \mathrm{dx}$ and ${e}^{x} - 1 = u - 2$. Then $\mathrm{dx} = \frac{1}{e} ^ \left(x\right) \mathrm{du} = \frac{1}{u - 2} \mathrm{du}$ and the integral becomes

$\setminus \int \frac{{e}^{x} - 1}{{e}^{x} + 1} \mathrm{dx} = \setminus \int \frac{u - 2}{u \left(u - 1\right)} \mathrm{du}$.

Next, use the Method of Partial Fractions to write $\frac{u - 2}{u \left(u - 1\right)}$ as $\frac{2}{u} - \frac{1}{u - 1}$, which can be easily integrated to get $2 \ln | u | - \ln | u - 1 | + C$.

Now substitute $u = {e}^{x} + 1$, which is never negative, and use the fact that $\ln \left({e}^{x}\right) = x$ to get

$\setminus \int \frac{{e}^{x} - 1}{{e}^{x} + 1} \mathrm{dx} = 2 \ln \left({e}^{x} + 1\right) - x + C$.

This can be checked by differentiation:

$\frac{d}{\mathrm{dx}} \left(2 \ln \left({e}^{x} + 1\right) - x + C\right) = \frac{2}{{e}^{x} + 1} \cdot {e}^{x} - 1 = \frac{2 {e}^{x} - \left({e}^{x} + 1\right)}{{e}^{x} + 1} = \frac{{e}^{x} - 1}{{e}^{x} + 1}$

Sep 12, 2017

$\int \frac{{e}^{x} - 1}{{e}^{x} + 1} \mathrm{dx} = 2 \ln \left({e}^{x} + 1\right) - x + C$

#### Explanation:

Given: $\int \frac{{e}^{x} - 1}{{e}^{x} + 1} \mathrm{dx}$

Insert zero into the numerator:

$\int \frac{{e}^{x} + 0 - 1}{{e}^{x} + 1} \mathrm{dx}$

In place of the 0 we write ${e}^{x} - {e}^{x}$:

$\int \frac{{e}^{x} + {e}^{x} - {e}^{x} - 1}{{e}^{x} + 1} \mathrm{dx}$

Now we do some clever grouping:

$\int \frac{\left({e}^{x} + {e}^{x}\right) - \left({e}^{x} + 1\right)}{{e}^{x} + 1} \mathrm{dx}$

We combine the first group:

$\int \frac{2 {e}^{x} - \left({e}^{x} + 1\right)}{{e}^{x} + 1} \mathrm{dx}$

Separate into two fractions:

$\int \frac{2 {e}^{x}}{{e}^{x} + 1} - \frac{{e}^{x} + 1}{{e}^{x} + 1} \mathrm{dx}$

Please notice that the second fraction becomes 1:

$\int \frac{2 {e}^{x}}{{e}^{x} + 1} - 1 \mathrm{dx}$

Separate into two integrals:

$2 \int {e}^{x} / \left({e}^{x} + 1\right) \mathrm{dx} - \int \mathrm{dx}$

For the first integral we let $u = {e}^{x} + 1$, then $\mathrm{du} = {e}^{x} \mathrm{dx}$:

$2 \int \frac{1}{u} \mathrm{du} - \int \mathrm{dx}$

We know these integrals very well:

$2 \ln | u | - x + C$

Reverse the substitution:

$2 \ln \left({e}^{x} + 1\right) - x + C$

Sep 12, 2017

$x + 2 \ln \left(1 + {e}^{-} x\right) + C$

#### Explanation:

We can also do this integral as follows:

$\int \frac{{e}^{x} - 1}{{e}^{x} + 1} \mathrm{dx} = \frac{{e}^{x} + 1 - 2}{{e}^{x} + 1} \mathrm{dx}$

Splitting up the fraction as $\frac{{e}^{x} + 1}{{e}^{x} + 1} - \frac{2}{{e}^{x} + 1}$, this becomes:

$= \int \left(1 - \frac{2}{{e}^{x} + 1}\right) \mathrm{dx}$

We can integrate the first term easily:

$= x - 2 \int \frac{1}{{e}^{x} + 1} \mathrm{dx}$

Now, multiply the integrand by ${e}^{-} \frac{x}{e} ^ - x$. This seems ridiculous, but you'll see why it works in a second:

$= x - 2 \int \frac{{e}^{-} x}{1 + {e}^{-} x} \mathrm{dx}$

Let $u = 1 + {e}^{-} x$, implying that $\mathrm{du} = - {e}^{-} x \mathrm{dx}$. Then:

$= x + 2 \int \frac{1}{u} \mathrm{du}$

$= x + 2 \ln \left\mid u \right\mid + C$

$= x + 2 \ln \left\mid 1 + {e}^{-} x \right\mid + C$

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

$x + 2 \ln \left(1 + {e}^{-} x\right) + C$

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