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

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To evaluate the integral
$$
\int \frac{\sqrt{64 - x^{2}}}{x} \, dx,
$$
we can use a trigonometric substitution. Here's a step-by-step solution:

### Step 1: Trigonometric Substitution
Let $$ x = 8 \sin \theta $$. Then:
$$
dx = 8 \cos \theta \, d\theta,
$$
and
$$
\sqrt{64 - x^2} = \sqrt{64 - 64 \sin^2 \theta} = 8 \cos \theta.
$$

### Step 2: Substitute into the Integral
Substituting $$ x = 8 \sin \theta $$ and $$ dx = 8 \cos \theta \, d\theta $$ into the integral:
$$
\int \frac{8 \cos \theta}{8 \sin \theta} \cdot 8 \cos \theta \, d\theta = \int \frac{\cos \theta}{\sin \theta} \cdot 8 \cos \theta \, d\theta = 8 \int \frac{\cos^2 \theta}{\sin \theta} \, d\theta.
$$

### Step 3: Simplify the Integrand
Using the identity $$ \cos^2 \theta = 1 - \sin^2 \theta $$:
$$
8 \int \frac{1 - \sin^2 \theta}{\sin \theta} \, d\theta = 8 \left( \int \csc \theta \, d\theta - \int \sin \theta \, d\theta \right).
$$

### Step 4: Integrate
We know:
$$
\int \csc \theta \, d\theta = -\ln \left| \csc \theta + \cot \theta \right| + C,
$$
and
$$
\int \sin \theta \, d\theta = -\cos \theta + C.
$$
Thus,
$$
8 \left( -\ln \left| \csc \theta + \cot \theta \right| + \cos \theta \right) + C.
$$

### Step 5: Back-Substitute $$ \theta $$ in Terms of $$ x $$
Recall that:
$$
\sin \theta = \frac{x}{8}, \quad \cos \theta = \frac{\sqrt{64 - x^2}}{8}.
$$
So,
$$
\csc \theta = \frac{8}{x}, \quad \cot \theta = \frac{\sqrt{64 - x^2}}{x}.
$$
Substituting these back into the expression:
$$
-8 \ln \left| \frac{8 + \sqrt{64 - x^2}}{x} \right| + \sqrt{64 - x^2} + C.
$$

### Final Answer
Combining all the steps, the integral evaluates to:
$$
\boxed{\,\sqrt{64 - x^{2}} \; -\; 8\,\ln\left(\frac{8 + \sqrt{64 - x^{2}}}{\,x\,}\right) + C\,}
$$
The integral evaluates to: $$ \sqrt{64 - x^{2}} - 8 \ln\left(\frac{8 + \sqrt{64 - x^{2}}}{x}\right) + C $$

Use integral tables to evaluate the integral.

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