How to evaluate the integral $\int\frac{\sqrt{x^2-9}}{x^3}\;\mathrm d x$? $$\int\frac{\sqrt{x^2-9}}{x^3}\;\mathrm d x$$
The question is ask me to evaluate the integral but I have no idea how to start?
If there are any formulas required for this question, can you please list them ?
Thank you for any help!

If you know the formula 
\begin{gather*}\tag{$\star$}
\int \sqrt{1-t^2}\,\rm{d}t=\frac{1}{2}\left(t\sqrt{1-t^2}+\arcsin(t)+C\right),
\end{gather*}
you can evaluate the integral as follows. 
Since 
\begin{align*}
&\quad \frac{\sqrt{x^2-9}}{x^3}\rm{d} x=\frac{\sqrt{x^2\big(1-(3/x)^2\big)}}{x^3}\rm{d} x\\
&=\sgn(x)\cdot\frac{x\sqrt{1-(3/x)^2}}{x^3}\rm{d} x=\sgn(x)\sqrt{1-(3/x)^2}\cdot\frac{1}{x^2}\rm{d} x\\
&=-\frac{\sgn(x)}{3}\sqrt{1-\left(\frac{3}{x}\right)^2}\rm{d}\left(\frac{3}{x}\right),
\end{align*}
together with $(\star)$ we have
\begin{align*}
\int\frac{\sqrt{x^2-9}}{x^3}\rm{d} x&=-\frac{\sgn(x)}{6}\left(\frac{3}{x}\sqrt{1-\left(\frac{3}{x}\right)^2}+\arcsin\left(\frac{3}{x}\right)\right)+C\\
&=-\frac{\sgn(x)}{2}\left(\frac{1}{x}\cdot\sqrt{1-\frac{9}{x^2}}+\frac{1}{3}\arcsin\left(\frac{3}{x}\right)\right)+C.
\end{align*}
Here $\sgn(x)$ is the signum function.
