The integral is:

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

To solve this, we use substitution. Let $v = 1 - e^u$. Then $e^u = 1 - v$, and $dv = -e^u \, du$.

Substituting into the integral:

1. Rewrite the numerator $e^u \, du = -dv$.
2. Rewrite the denominator $(1 - e^u)^2 = v^2$.

The integral becomes:

$\int \frac{-dv}{v^2}$

This simplifies to:

$-\int v^{-2} \, dv$

Now, integrate $v^{-2}$:

$-\int v^{-2} \, dv = -\left( -\frac{1}{v} \right) = \frac{1}{v}$

Substitute back $v = 1 - e^u$:

$\frac{1}{v} = \frac{1}{1 - e^u}$

So, the final result is:

$\frac{1}{1 - e^u} + C$

where $C$ is the constant of integration.

Would you like further clarification or a detailed breakdown of any step? Here are some related questions for practice:

1. How do you determine when substitution is the best approach for solving an integral?
2. How would you solve $\int \frac{e^u}{1 - e^u} \, du$?
3. What happens if the substitution doesn't simplify the integral sufficiently?
4. Can this integral be solved using a different method, such as partial fractions?
5. How would the solution change if the denominator were $(1 - e^u)^n$, where $n \neq 2$?

Tip: Always check if a substitution simplifies both the numerator and denominator—it often leads to an easier integral.