Answer
$$\frac{1}{3}\sec^{-1}\left(\frac{x}{3}\right)+C$$
Work Step by Step
Given
$$\int \frac{d x}{x \sqrt{x^{2}-9}} $$
Let $$x=3\sec \theta \to dx= 3\sec \theta \tan \theta d \theta $$
Then
\begin{aligned}
\int \frac{d x}{x \sqrt{x^{2}-9}} &=\int \frac{3 \sec \theta \tan \theta d \theta}{3 \sec \theta \sqrt{(3 \sec \theta)^{2}-9}} \\
&=\int \frac{3 \sec \theta \cdot \tan \theta d \theta}{3 \sec \theta \sqrt{9 \sec ^{2} \theta-9}} \\
&=\int \frac{\tan \theta d \theta}{3 \sqrt{\sec ^{2} \theta-1}} \\
&=\int \frac{\tan \theta d \theta}{3 \tan ^{2} \theta-1} \\
&=\int \frac{\tan \theta d \theta}{3 \tan ^{2} \theta} \\
&=\int \frac{\tan \theta d \theta}{3 \tan \theta} \\
&=\frac{1}{3} \int d \theta \\
&=\frac{1}{3} \theta+C\\
&=\frac{1}{3}\sec^{-1}\left(\frac{x}{3}\right)+C
\end{aligned}
