# How do you find the antiderivative of  {(e^x)/ [(e^(2x)) - 1]}?

Oct 22, 2016

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

#### Explanation:

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

Let $u = {e}^{x}$ so that $\mathrm{du} = {e}^{x} \mathrm{dx}$:

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

Now, we can ley $u = \sec \theta$. This implies that $\mathrm{du} = \sec \theta \tan \theta d \theta$.

Plugging these in:

$I = \int \frac{\sec \theta \tan \theta}{{\sec}^{2} \theta - 1} d \theta$

Note that $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$, so ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$:

$I = \int \frac{\sec \theta \tan \theta}{\tan} ^ 2 \theta d \theta$

$I = \int \sec \frac{\theta}{\tan} \theta d \theta$

$I = \int \frac{1}{\cos} \theta \left(\cos \frac{\theta}{\sin} \theta\right) d \theta$

$I = \int \csc \theta$

This is a fairly common integral:

$I = - \ln \left(\left\mid \csc \theta + \cot \theta \right\mid\right)$

We need to rewrite this using $u = \sec \theta$. This means we have a right triangle where $u$ is the hypotenuse, $1$ is the side adjacent to $\theta$, and $\sqrt{{u}^{2} - 1}$ is the side opposite $\theta$.

Thus $\csc \theta = \frac{u}{\sqrt{{u}^{2} - 1}}$ and $\cot \theta = \frac{1}{\sqrt{{u}^{2} - 1}}$.

So:

$I = - \ln \left(\left\mid \frac{u}{\sqrt{{u}^{2} - 1}} + \frac{1}{\sqrt{{u}^{2} - 1}} \right\mid\right) + C$

$I = - \ln \left(\left\mid \frac{u + 1}{\sqrt{{u}^{2} - 1}} \right\mid\right) + C$

Bringing in the negative $1$ as an exponent through the rule $B \log \left(A\right) = \log \left({A}^{B}\right)$:

$I = \ln \left(\left\mid \frac{\sqrt{{u}^{2} - 1}}{u + 1} \right\mid\right) + C$

Through $u = {e}^{x}$:

$I = \ln \left(\left\mid \frac{\sqrt{{e}^{2 x} - 1}}{{e}^{x} + 1} \right\mid\right) + C$

Notice that $\sqrt{{e}^{2 x} - 1} > 0$ and ${e}^{x} + 1 > 0$ for all values of $x$, so the absolute value bars can be removed:

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