Answer
$$ 8\left( \frac{x}{4}\frac{\sqrt{16-x^2}}{4}- \ln | \frac{x}{4}+\frac{\sqrt{16-x^2}}{4}| \right)+C$$
Work Step by Step
Given $$ \int \sqrt{x^{2}-16} d x$$
Let
\begin{align*}
x&=4\sec u\\
dx&=4\sec u\tan udu
\end{align*}
Then
\begin{align*}
\int \sqrt{x^{2}-16} d x&= 4\int \sqrt{16\sec^2 u-16} \sec u\tan udu\\
&=4\int \sqrt{16\tan^2 u-} \sec u\tan udu\\
&=16\int \sec u\tan^2 udu\\
&=16\int (\sec^3 u-\sec u )du,\ \ \\
& \text{From the table: }\\
&= 16\left(\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|-\ln |\sec u+\tan u| \right)+C\\
&= 16\left(\frac{1}{2} \sec u \tan u-\frac{1}{2} \ln |\sec u+\tan u| \right)+C\\
&= 8\left( \frac{x}{4}\frac{\sqrt{16-x^2}}{4}- \ln | \frac{x}{4}+\frac{\sqrt{16-x^2}}{4}| \right)+C
\end{align*}