Which of the following is equivalent to $\int \frac{1}{x^{2}-16} d x$ ?
(A) $\int \frac{1}{u} d u$, where $u=x^{2}-16$
(B) $\frac{1}{2 x} \int \frac{1}{u} d u$, where $u=x^{2}-16$
(C) $\frac{1}{8} \int \frac{1}{x-4} d x-\frac{1}{8} \int \frac{1}{x+4} d x$
(D) $\int \frac{1}{x-4} d x+\int \frac{1}{x+4} d x$

Step 1 ：The integral can be rewritten as \( \int \frac{1}{(x-4)(x+4)} dx \).

Step 2 ：This can be decomposed into two simpler fractions \( \frac{A}{x-4} + \frac{B}{x+4} \), where A and B are constants to be determined.

Step 3 ：We can find A and B by equating the original fraction with the decomposed fraction and solving for A and B.

Step 4 ：The equation obtained by equating the original fraction with the decomposed fraction is \( \frac{1}{x^2 - 16} = \frac{A}{x - 4} + \frac{B}{x + 4} \).

Step 5 ：Solving this equation gives the solution \( A = \frac{1}{8} \) and \( B = -\frac{1}{8} \).

Step 6 ：Therefore, the integral can be rewritten as \( \frac{1}{8} \int \frac{1}{x-4} dx - \frac{1}{8} \int \frac{1}{x+4} dx \).

Step 7 ：This matches with option (C). So, option (C) is the correct answer.

Step 8 ：Final Answer: \(\boxed{(C) \frac{1}{8} \int \frac{1}{x-4} dx - \frac{1}{8} \int \frac{1}{x+4} dx}\)