Evaluate the integral of (e^x + e^-x) / (e^x - e^-x) dx from 0 to infinity.

The integral to evaluate is:

$$\int_{0}^{\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}} \,dx$$

The integrand can be recognized as the hyperbolic tangent function, $\tanh x$.
Recall that $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$.
Therefore, $\frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{2 \sinh x}{2 \cosh x} = \tanh x$.
The integral becomes:

$$\int_{0}^{\infty} \tanh x \,dx$$

We know that $\tanh x = \frac{\sinh x}{\cosh x}$.
Let $u = \cosh x$.
Then, $du = \sinh x \,dx$.
Substituting these into the integral:

$$\int \frac{1}{u} \,du = \ln|u| + C$$

Since $\cosh x = \frac{e^x + e^{-x}}{2}$ is always positive for real $x$, we can write:

$$\int \tanh x \,dx = \ln(\cosh x) + C$$

Since the upper limit is $\infty$, this is an improper integral. We evaluate it using a limit:

$$\int_{0}^{\infty} \tanh x \,dx = \lim_{b \to \infty} \int_{0}^{b} \tanh x \,dx$$

Substitute the indefinite integral:

$$= \lim_{b \to \infty} \left[ \ln(\cosh x) \right]_{0}^{b}$$

Apply the limits of integration:

$$= \lim_{b \to \infty} (\ln(\cosh b) - \ln(\cosh 0))$$

First, evaluate $\cosh 0$:

$$\cosh 0 = \frac{e^0 + e^{-0}}{2} = \frac{1+1}{2} = 1$$

So, $\ln(\cosh 0) = \ln(1) = 0$.

Next, evaluate the limit of $\ln(\cosh b)$ as $b \to \infty$:
As $b \to \infty$, $e^b \to \infty$ and $e^{-b} \to 0$.
Therefore, $\cosh b = \frac{e^b + e^{-b}}{2} \to \infty$ as $b \to \infty$.
Since $\ln(y) \to \infty$ as $y \to \infty$, we have:

$$\lim_{b \to \infty} \ln(\cosh b) = \infty$$

Substitute these values back into the expression from Step 3:

$$\int_{0}^{\infty} \tanh x \,dx = \infty - 0 = \infty$$

Since the limit is infinite, the integral diverges.

The final answer is $\text{Diverges}$.