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