We are tasked with evaluating the integral:

$I = \int \frac{-4}{x^2 \sqrt{16x^2 - 9}} \, dx$

To solve this, we will use a substitution method. Let's follow the steps:

Step 1: Simplifying the square root expression

We notice that $\sqrt{16x^2 - 9}$ resembles a form that can benefit from a trigonometric substitution. Let’s try to simplify this expression first.

• Factor the term inside the square root:

$\sqrt{16x^2 - 9} = \sqrt{16\left(x^2 - \frac{9}{16}\right)}$

This leads us to consider a substitution of the form $x = \frac{3}{4} \sec(\theta)$ because it matches the structure of $a^2 \sec^2(\theta) - b^2$. After applying this, we can proceed to solve the integral.

Let me calculate the result step by step.The integral evaluates to a piecewise result:

\begin{cases} \frac{-4i \sqrt{-16 + \frac{9}{x^2}}}{9}, & \text{if} \ \frac{9}{x^2} > 16 \\ \frac{-4 \sqrt{16 - \frac{9}{x^2}}}{9}, & \text{otherwise} \end{cases}$$ This complex result suggests that the integral may not have a simple real solution over all values of $$x$$ without considering conditions on the domain. Depending on the value of $$x$$, the square root can result in complex values when the expression inside becomes negative. Let me know if you would like further explanation of any steps! ### Follow-up Questions: 1. What conditions can ensure the integral remains real? 2. How would a trigonometric substitution simplify the square root? 3. What role does the imaginary unit $$i$$ play in the result? 4. How would the answer change if the domain of $$x$$ was restricted? 5. Can we interpret the physical significance of complex results in an applied problem? **Tip**: Always check the domain of the function inside square roots when integrating to avoid complex results where real values are expected!