Find the indefinite integral. (Use $$ C $$ for the constant of integration.) $$ \int \frac{\sqrt{x^{2}-16}}{x} d x $$

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We start with the integral
$$
\int \frac{\sqrt{x^2-16}}{x}\,dx.
$$
A common strategy for integrals involving $$\sqrt{x^2-16}$$ is to use a trigonometric substitution. Let
$$
x = 4\sec\theta.
$$
Then,
$$
\sqrt{x^2-16} = \sqrt{16\sec^2\theta-16} = \sqrt{16(\sec^2\theta-1)} = 4\tan\theta,
$$
and the differential becomes
$$
dx = 4\sec\theta\tan\theta\,d\theta.
$$

Substitute these into the integral:
$$
\int \frac{4\tan\theta}{4\sec\theta} \cdot \left(4\sec\theta\tan\theta\,d\theta\right).
$$
Simplify step by step:
- The factor $$\frac{4\tan\theta}{4\sec\theta}$$ reduces to $$\frac{\tan\theta}{\sec\theta}$$.
- Multiplying by $$4\sec\theta\tan\theta\,d\theta$$ gives:
$$
\frac{\tan\theta}{\sec\theta} \cdot 4\sec\theta\tan\theta\,d\theta = 4\tan^2\theta\,d\theta.
$$
Thus, the integral becomes:
$$
4\int \tan^2\theta\,d\theta.
$$

Recall that $$\tan^2\theta = \sec^2\theta - 1$$. Therefore,
$$
4\int \tan^2\theta\,d\theta = 4\int (\sec^2\theta-1)\,d\theta.
$$
Now integrate term by term:
$$
4\left(\int \sec^2\theta\,d\theta - \int d\theta\right) = 4\left(\tan\theta - \theta\right)+C.
$$

Next, we express the answer in terms of $$x$$. We have:
$$
x = 4\sec\theta \quad \Longrightarrow \quad \sec\theta = \frac{x}{4},
$$
and
$$
\tan\theta = \sqrt{\sec^2\theta-1} = \sqrt{\left(\frac{x}{4}\right)^2-1} = \sqrt{\frac{x^2}{16}-1} = \frac{\sqrt{x^2-16}}{4}.
$$
Thus,
$$
4\left(\tan\theta - \theta\right) = 4\left(\frac{\sqrt{x^2-16}}{4} - \theta\right) = \sqrt{x^2-16} - 4\theta.
$$

Finally, recall that
$$
\theta = \sec^{-1}\left(\frac{x}{4}\right).
$$
Substituting back, the antiderivative is
$$
\sqrt{x^2-16} - 4\sec^{-1}\left(\frac{x}{4}\right) + C.
$$

Therefore, the indefinite integral is
$$
\boxed{\sqrt{x^2-16} - 4\sec^{-1}\left(\frac{x}{4}\right) + C.}
$$
The indefinite integral is $$ \sqrt{x^2-16} - 4\sec^{-1}\left(\frac{x}{4}\right) + C $$.

Find the indefinite integral. (Use for the constant of integration.)

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