Use integral tables to evaluate the integral. \[ \int \frac{\sqrt{64-x^{2}}}{x} d x \]

To evaluate the integral \(\int \frac{\sqrt{64-x^{2}}}{x} \, dx\), we can use a trigonometric substitution. Since we have \(\sqrt{64 - x^2}\), we can let \(x = 8\sin(\theta)\). This implies \(dx = 8\cos(\theta) \, d\theta\) and \(\sqrt{64 - x^2} = \sqrt{64(1 - \sin^2(\theta))} = 8\cos(\theta)\). Now we can transform the integral: \[ \int \frac{\sqrt{64-x^{2}}}{x} \, dx = \int \frac{8\cos(\theta)}{8\sin(\theta)} \cdot (8\cos(\theta) \, d\theta) = \int \frac{8\cos^2(\theta)}{\sin(\theta)} \, d\theta = 8 \int \frac{\cos^2(\theta)}{\sin(\theta)} \, d\theta. \] We can rewrite \(\cos^2(\theta)\) as \(\frac{1 - \sin^2(\theta)}{2}\): \[ 8 \int \frac{\cos^2(\theta)}{\sin(\theta)} \, d\theta = 8 \int \frac{1 - \sin^2(\theta)}{2\sin(\theta)} \, d\theta = 4 \int \frac{1}{\sin(\theta)} \, d\theta - 4 \int \sin(\theta) \, d\theta. \] The first integral is: \[ 4 \int \csc(\theta) \, d\theta = 4 \ln | \csc(\theta) - \cot(\theta) | + C, \] and the second integral is: \[ -4 (-\cos(\theta)) = 4\cos(\theta). \] Putting it all together, we have: \[ 4 \ln | \csc(\theta) - \cot(\theta) | + 4\cos(\theta) + C. \] Now we need to convert everything back in terms of \(x\). We have: \(\sin(\theta) = \frac{x}{8}\) and \(\cos(\theta) = \sqrt{1 - \left(\frac{x}{8}\right)^2} = \frac{\sqrt{64 - x^2}}{8}\). Also, \(\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{8}{x}\) and \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{\sqrt{64 - x^2}}{x}\). Thus, the expression becomes: \[ \int \frac{\sqrt{64-x^{2}}}{x} \, dx = 4 \ln \left| \frac{8}{x} - \frac{\sqrt{64-x^2}}{x} \right| + \sqrt{64 - x^2} + C. \] This simplifies to: \[ \int \frac{\sqrt{64-x^{2}}}{x} \, dx = 4 \ln \left| \frac{8 - \sqrt{64-x^2}}{x} \right| + \sqrt{64 - x^2} + C. \]