$\int \sqrt{x^2-6 x-16} \mathrm{~d} x$ equals.

Evaluation of the Integral

To evaluate $\int \sqrt{x^2-6x-16} \mathrm{d}x$, complete the square for the quadratic expression: $x^2 - 6x - 16 = (x-3)^2 - 25 = (x-3)^2 - 5^2$.

The integral becomes $\int \sqrt{(x-3)^2 - 5^2} \mathrm{d}x$, which matches the standard form $\int \sqrt{u^2 - a^2} \mathrm{d}u$ with $u = x-3$ and $a = 5$.

Using the standard formula:
$\int \sqrt{u^2 - a^2} \mathrm{d}u = \frac{u}{2} \sqrt{u^2 - a^2} - \frac{a^2}{2} \log |u + \sqrt{u^2 - a^2}| + C$.

Substitute $u = x-3$ and $a = 5$:
$= \frac{x-3}{2} \sqrt{(x-3)^2 - 5^2} - \frac{25}{2} \log |(x-3) + \sqrt{(x-3)^2 - 5^2}| + C$.

Simplify to:
$= \frac{x-3}{2} \sqrt{x^2 - 6x - 16} - \frac{25}{2} \log |x-3 + \sqrt{x^2 - 6x - 16}| + C$.

This matches Option B.

Final Answer: $\boxed{\text{B}}$