\( \int \frac{d x}{x \sqrt{x^{2}-16}} \) ? (Hint: Scroll to either the header or the footer to see the formula page.) \( \frac{1}{4} \cos ^{-1}\left(\frac{4}{x}\right)+C \) \( \frac{1}{4} \sin ^{-1}\left(\frac{4}{x}\right)+C \) \( \frac{1}{4} \operatorname{sen}^{-1}\left(\frac{4}{x}\right)+C \) None of these responses is correct. Nest \( \left.\frac{4}{x}\right)+C \)

We start with the integral \[ I=\int \frac{dx}{x\sqrt{x^2-16}}. \] **Step 1. Substitute \( x=4\sec\theta \):** Let \[ x = 4\sec\theta. \] Then, the derivative is \[ dx = 4\sec\theta\,\tan\theta\,d\theta. \] Also, note that \[ x^2-16 = 16\sec^2\theta - 16 = 16(\sec^2\theta-1) = 16\tan^2\theta. \] Thus, \[ \sqrt{x^2-16} = 4|\tan\theta|. \] Assuming the range of \(\theta\) where \(\tan\theta \ge 0\) (this can be arranged by choosing an appropriate interval), we have \[ \sqrt{x^2-16} = 4\tan\theta. \] **Step 2. Substitute into the integral:** Replace \(x\), \(dx\), and \(\sqrt{x^2-16}\) in the original integral: \[ I=\int \frac{4\sec\theta\tan\theta\,d\theta}{4\sec\theta\,(4\tan\theta)}. \] Simplify the expression: \[ I=\int \frac{4\sec\theta\tan\theta\,d\theta}{16\sec\theta\tan\theta} = \frac{1}{4} \int d\theta. \] Thus, \[ I=\frac{1}{4}\theta+C. \] **Step 3. Express \(\theta\) in terms of \(x\):** Recall that \[ x=4\sec\theta \quad \Longrightarrow \quad \sec\theta=\frac{x}{4}. \] Since \[ \theta = \sec^{-1}\left(\frac{x}{4}\right), \] and using the identity \[ \sec^{-1}\left(\frac{x}{4}\right)=\cos^{-1}\left(\frac{4}{x}\right), \] we have \[ \theta = \cos^{-1}\left(\frac{4}{x}\right). \] **Step 4. Write the final answer:** Substitute back into the expression for \( I \): \[ I=\frac{1}{4}\cos^{-1}\left(\frac{4}{x}\right)+C. \] Thus, the correct answer is \[ \frac{1}{4}\cos^{-1}\left(\frac{4}{x}\right)+C. \]