The integral you've provided is:

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

Solution:

We can simplify this expression by breaking it into simpler components.

1. Rewrite the integrand: The given expression can be split as follows:

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

   So, the integral becomes:

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

2. First part: The integral of $1$ is straightforward:

   $\int 1 \, dx = x$

3. Second part: The integral $\int \frac{1}{e^x - 1} \, dx$ can be expressed in terms of the logarithm of the hyperbolic function. This is a standard integral, and its result is:

   $\int \frac{1}{e^x - 1} \, dx = \ln |e^x - 1| + C$

4. Combining results: Now, combining both parts, we get:

   $\int \frac{e^x + 1}{e^x - 1} \, dx = x + 2 \ln |e^x - 1| + C$

Thus, the final answer is:

$x + 2 \ln |e^x - 1| + C$

Would you like further details or clarifications?

Relative questions:

1. How can you solve integrals involving exponential functions more easily?
2. What are some common hyperbolic integrals and their solutions?
3. Why is logarithmic substitution helpful in certain integrals?
4. How do we handle absolute values when integrating?
5. What role does the constant $C$ play in indefinite integrals?

Tip:

Always consider splitting fractions and breaking complex expressions into simpler parts for easier integration.