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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.

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

Learn how to solve the integral \( \int \frac{e^u}{(1 - e^u)^2} \, du \) with a detailed, step-by-step substitution method. This guide covers key concepts in integration and is suitable for college-level calculus students.

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