To solve the integral $\int \frac{3}{e^{3x} - e^x} \, dx$, let's go step by step.

Step 1: Simplify the denominator

The expression in the denominator is $e^{3x} - e^x$. We can factor this as:

$e^{3x} - e^x = e^x(e^{2x} - 1)$

Step 2: Rewrite the integral

The integral now becomes:

$\int \frac{3}{e^x(e^{2x} - 1)} \, dx$

Step 3: Substitution

Let $u = e^x$, then $du = e^x \, dx$. This simplifies the integral as follows:

$\int \frac{3}{u(u^2 - 1)} \, du$

Now, we can simplify this using partial fraction decomposition.

Step 4: Partial fraction decomposition

We decompose the fraction:

$\frac{3}{u(u^2 - 1)} = \frac{A}{u} + \frac{B}{u-1} + \frac{C}{u+1}$

Multiplying through by $u(u^2 - 1) = u(u-1)(u+1)$, we get:

$3 = A(u-1)(u+1) + B u(u+1) + C u(u-1)$

Expanding both sides and equating coefficients gives values for $A$, $B$, and $C$.

Step 5: Solve for A, B, C

The solution yields:

$\frac{3}{u(u^2 - 1)} = \frac{1}{u} - \frac{1}{u-1} - \frac{1}{u+1}$

Step 6: Integrate each term

Now we can integrate term by term:

$\int \left( \frac{1}{u} - \frac{1}{u-1} - \frac{1}{u+1} \right) du$

This gives:

$\ln|u| - \ln|u-1| - \ln|u+1| + C$

Step 7: Back-substitute $u = e^x$

Substitute $u = e^x$ back into the equation:

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

Simplifying further:

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

Thus, the solution to the integral is:

$\boxed{x - \ln|e^x - 1| - \ln|e^x + 1| + C}$

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Would you like more details on any part of the process?

Here are some related questions:

1. How does partial fraction decomposition work in detail?
2. Why did we choose $u = e^x$ for substitution?
3. How can we simplify logarithmic expressions involving $e^x$?
4. What is the method of integrating rational functions?
5. How can we use substitution more effectively in other integrals?

Tip: When integrating fractions involving exponentials, substitution often simplifies the problem drastically.