The integral to solve is:

\[
\int \frac{1}{u^2 - 16} \, du
\]

To solve this, we can rewrite the denominator as a difference of squares:

\[
u^2 - 16 = (u - 4)(u + 4)
\]

Now, we'll use **partial fraction decomposition**. We express the integrand as:

\[
\frac{1}{(u - 4)(u + 4)} = \frac{A}{u - 4} + \frac{B}{u + 4}
\]

Multiplying both sides by \( (u - 4)(u + 4) \) gives:

\[
1 = A(u + 4) + B(u - 4)
\]

Expanding the right-hand side:

\[
1 = A(u + 4) + B(u - 4) = A u + 4A + B u - 4B
\]

Now, group the terms involving \( u \):

\[
1 = (A + B)u + (4A - 4B)
\]

This gives two equations:

1. \( A + B = 0 \)
2. \( 4A - 4B = 1 \)

From the first equation, we have \( A = -B \). Substituting this into the second equation:

\[
4A - 4(-A) = 1 \quad \Rightarrow \quad 4A + 4A = 1 \quad \Rightarrow \quad 8A = 1 \quad \Rightarrow \quad A = \frac{1}{8}
\]

Since \( A = \frac{1}{8} \), we have \( B = -\frac{1}{8} \).

Now, we can rewrite the integral as:

\[
\int \frac{1}{(u - 4)(u + 4)} \, du = \int \left( \frac{1/8}{u - 4} - \frac{1/8}{u + 4} \right) \, du
\]

We can split the integral:

\[
= \frac{1}{8} \int \frac{1}{u - 4} \, du - \frac{1}{8} \int \frac{1}{u + 4} \, du
\]

The integrals are standard logarithmic forms:

\[
= \frac{1}{8} \ln |u - 4| - \frac{1}{8} \ln |u + 4| + C
\]

Using properties of logarithms:

\[
= \frac{1}{8} \ln \left( \frac{|u - 4|}{|u + 4|} \right) + C
\]

Thus, the solution to the integral is:

\[
\int \frac{1}{u^2 - 16} \, du = \frac{1}{8} \ln \left( \frac{|u - 4|}{|u + 4|} \right) + C
\]

---

Would you like further clarification or details on any step?

Here are 5 related questions:

1. How does partial fraction decomposition work for more complex denominators?
2. How do you integrate functions with quadratic denominators?
3. What are other methods for integrating rational functions?
4. How can you use substitution to simplify integrals like this one?
5. How can logarithmic properties help simplify integral results?

**Tip**: Always check if a denominator can be factored, as it often simplifies the integral significantly.

Learn how to solve the integral of 1/(u^2 - 16) step by step using partial fraction decomposition. This detailed solution covers key concepts in calculus, including the difference of squares, logarithmic integration, and properties of logarithms. Suitable for students in undergraduate calculus.

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