Answer
$$\frac{1}{4} \ln \left|\frac{\sqrt{x^{2}+16}-4}{x}\right|+C$$
Work Step by Step
Given $$\int \frac{d x}{x \sqrt{x^{2}+16}} $$
Let $$x= 4\tan \theta ,\ \ \ dx= 4\sec^2\theta d\theta $$
Then
\begin{aligned} \int \frac{d x}{x \sqrt{x^{2}+16}} &=\int \frac{4 \sec ^{2} \theta d \theta}{4 \tan \theta \sqrt{(4 \tan \theta)^{2}+16}} \\ &=\int \frac{A \sec ^{2} \theta d \theta}{A \tan \theta \sqrt{16 \tan ^{2} \theta+16}} \\ &=\int \frac{\sec ^{2} \theta d \theta}{4 \tan \theta \sqrt{\sec ^{2} \theta}} \\ &=\frac{1}{4} \int \frac{\sec ^{2} \theta d \theta}{\tan \theta \sec \theta} \\ &=\frac{1}{4} \int \frac{\sec \theta d \theta}{\tan \theta} \\
&=\frac{1}{4} \int \csc \theta d \theta \\
&= \frac{1}{4}\ln | \csc \theta -\cos\theta |\\
&= \frac{1}{4} \ln \left|\frac{\sqrt{x^{2}+16}-4}{x}\right|+C
\end{aligned}
