Answer
$$\int \frac{\sqrt{16-x^{2}}}{x} d x =\ln |\frac{4}{x} +\frac{\sqrt{16-x^2 }}{x}|+\frac{\sqrt{16-x^2 }}{4}+C$$
Work Step by Step
$$
\int \frac{\sqrt{16-x^{2}}}{x} d x
$$
Let $ x=4 \sin \theta\ \ \ \to\ \ \ dx= 4\cos \theta d\theta $, then
\begin{align*}
\int \frac{\sqrt{16-x^{2}}}{x} d x&=\int \frac{\sqrt{16-16\sin^{2}}\theta }{4 \sin \theta} 4\cos \theta d\theta\\
&=\int \frac{4\cos^2 \theta d\theta}{4 \sin \theta} \\
&=\int \frac{( 1- \sin^2 \theta )d\theta}{ \sin \theta}\\
&=\int(\csc\theta -\sin\theta)d\theta \\
&=\ln |\csc\theta +\cot\theta|+\cos\theta+C\\
&=\ln |\frac{4}{x} +\frac{\sqrt{16-x^2 }}{x}|+\frac{\sqrt{16-x^2 }}{4}+C
\end{align*}
