The given mathematical expression is:

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

Let me solve this step-by-step:

Step 1: Simplify the denominator

The denominator $e^{2x} - 1$ can be rewritten using exponential properties.

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

Thus, the integral becomes:

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

Step 2: Use partial fraction decomposition

We decompose the fraction:

$\frac{e^x}{\left(e^x - 1\right)\left(e^x + 1\right)} = \frac{A}{e^x - 1} + \frac{B}{e^x + 1}$

Multiply through by the denominator $\left(e^x - 1\right)\left(e^x + 1\right)$ to solve for $A$ and $B$:

$e^x = A\left(e^x + 1\right) + B\left(e^x - 1\right)$

Simplify:

$e^x = A e^x + A + B e^x - B$

$e^x = (A + B)e^x + (A - B)$

Equating coefficients of $e^x$ and the constant terms:

1. Coefficient of $e^x$: $A + B = 1$
2. Constant term: $A - B = 0$

Solve these equations:

1. $A + B = 1$
2. $A - B = 0 \implies A = B$

Substitute $A = B$ into the first equation:

$A + A = 1 \implies A = \frac{1}{2}, \, B = \frac{1}{2}$

Thus:

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

Step 3: Integrate each term

The integral now becomes:

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

First integral: $\int \frac{1}{e^x - 1} \, dx$

Let $u = e^x - 1 \implies du = e^x dx$. Then:

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

Second integral: $\int \frac{1}{e^x + 1} \, dx$

Let $v = e^x + 1 \implies dv = e^x dx$. Then:

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

Step 4: Combine the results

Now combine the results of both integrals:

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

This can be written as:

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

Final Answer:

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

Would you like additional clarifications or details?

Follow-up Questions:

1. How can partial fraction decomposition be applied to other rational functions?
2. What is the significance of logarithmic terms in integration results?
3. Can this integral be evaluated using substitution directly without partial fractions?
4. How does the behavior of the function $e^x/(e^{2x} - 1)$ change for large $x$?
5. Can this method be extended to integrals involving higher powers of exponentials?

Tip:

When dealing with integrals involving exponential terms, always check if substitution or partial fractions can simplify the problem before proceeding with integration.