Answer
$$\ln \left|\frac{x}{6}+\frac{\sqrt{x^{2}-36}}{6}\right|+C$$
Work Step by Step
Given $$\int \frac{d x}{\sqrt{x^{2}-36}}$$
Let
$$ x=6\sec u \ \ \ \ \ \ dx=6\sec u\tan udu $$
Then
\begin{aligned}
\int \frac{d x}{\sqrt{x^{2}-36}}&=\int \frac{1}{\sqrt{36 \sec ^{2} u-36}} 6 \sec u \tan u \, d u\\
&=\int \frac{1}{6 \sqrt{\sec ^{2} u-1}} 6 \sec u \tan u \, d u\\
&=\int \frac{1}{\tan u} \sec u \tan u \, d u\\
&=\int \sec u \, d u\\
&=\ln|\sec u+\tan u|+C\\
&= \ln \left|\frac{x}{6}+\frac{\sqrt{x^{2}-36}}{6}\right|+C
\end{aligned}
