To find the integral
[tex]\[
\int \frac{1}{x \sqrt{16 x^2-1}} \, dx,
\][/tex]
we will consider the result provided.
1. Identify the form of the result: From the result, we recognize that the integral can be represented using a piecewise function to capture different scenarios based on the constraints.
2. Express the integral result using piecewise notation:
[tex]\[
\int \frac{1}{x \sqrt{16 x^2-1}} \, dx = \begin{cases}
I \cdot \text{acosh}\left(\frac{1}{4x}\right) & \text{if} \ \frac{1}{|x^2|} > 16 \\
-\arcsin\left(\frac{1}{4x}\right) & \text{otherwise}
\end{cases} + C
\][/tex]
3. Interpret each case:
- The first case corresponds to complex values when [tex]\(\frac{1}{x^2} > 16\)[/tex]. It involves the inverse hyperbolic cosine function (`acosh`), indicating this scenario typically happens outside the typical range of real numbers.
- The second case entails more common scenarios, providing a result in terms of the inverse sine function (`arcsin`).
Therefore, the final piecewise solution for the integral is:
[tex]\[
\int \frac{1}{x \sqrt{16 x^2-1}} \, dx = \begin{cases}
I \cdot \text{acosh}\left(\frac{1}{4x}\right) & \text{if} \ \frac{1}{|x^2|} > 16 \\
-\arcsin\left(\frac{1}{4x}\right) & \text{otherwise}
\end{cases} + C,
\][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
