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!
 A: You want a substitution which will eliminate the square root, in other words
$$x^2-9=(\hbox{something})^2\ .$$
As we will see later, the $9$ is easy to handle, so we'll temporarily replace it by $1$ and consider
$$x^2-1=(\hbox{something})^2\ .$$
This should remind you of various formulae such as
$$\sec^2\theta-1=\tan^2\theta\ ,\quad\cosh^2\theta-1=\sinh^2\theta\ ;$$
perhaps you know some others too.  So two suggested substitutions are
$$x=3\sec\theta\ ,\quad x=3\cosh\theta\ ;$$
notice how the $3$ takes care of the $9$ which we ignored earlier.
There is no immediate way to tell which, if either, of these substitutions will work; you just have to try them and see what happens.
Good luck!
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\iff}{\Longleftrightarrow}
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\begin{align}
&\overbrace{\color{#66f}{\large\int{\root{x^{2} - 9} \over x^{3}}\,\dd x}}
^{\ds{\color{#c00000}{x \equiv {1 \over t}\ \imp\ t = {1 \over x}}}}\ =\
\int{\root{1/t^{2} - 9} \over 1/t^{3}}\,\pars{-\,{\dd t \over t^{2}}}
=-\ \overbrace{\int\root{1 - 9t^{2}}\,\dd t}
^{{\ds{\color{#c00000}{t \equiv {\sin\pars{\theta} \over 3}}}}}
\\[5mm]&=-\int\cos\pars{\theta}\,{\cos\pars{\theta} \over 3}\,\dd\theta
=-\,{1 \over 6}\int\bracks{1 + \cos\pars{2\theta}}\,\dd\theta
=-\,{1 \over 6}\,\theta - {1 \over 12}\,\sin\pars{2\theta}
\\[5mm]&=-\,{1 \over 6}
\bracks{\theta + \sin\pars{\theta}\root{1 - \sin^{2}\pars{\theta}}}
=-\,{1 \over 6}\bracks{\arcsin\pars{3t} + 3t\root{1 - 9t^{2}}}
\\[5mm]&=-\,{1 \over 6}\bracks{\arcsin\pars{3 \over x} + {3 \over x}
\root{1 - 9\,{1 \over x^{2}}}}
\end{align}

\begin{align}
&\color{#66f}{\large\int{\root{x^{2} - 9} \over x^{3}}\,\dd x}
=\color{#66f}{\large -\,{1 \over 6}\arcsin\pars{3 \over x}
-{\root{x^{2} - 9} \over 2x^{2}}} + \mbox{a constant}
\end{align}

A: HINT:
$$\int\frac{\sqrt{x^2-9}}{x(x^2)}dx =\int\frac{\sqrt{x^2-9}}{(x^2)^2}xdx$$
Write $\sqrt{x^2-9}=u\implies x^2-9=u^2$
A: Try this
$$ t^2 = x^2 - 9, \quad t\,dt = x\,dx $$
so
$$ \int \frac{\sqrt{x^2-9}}{x^3}\,dx = \int \frac{\sqrt{x^2-9}}{x^4} x\,dx \\
= \int \frac{t}{(t^2+9)^2} t\,dt \\= \int t \frac{t}{(t^2+9)^2} dt $$
Now do integration by parts
$$ u = t, \quad dv = \frac{t}{(t^2+9)^2}\,dt $$
$$ du = dt, \quad v = -\frac{1}{2(t^2+9)}$$
$$ \int \frac{t^2}{(t^2+9)^2} dt = -\frac{t}{2(t^2+9)} + \int \frac{1}{2(t^2+9)}\,dt \\
= -\frac{t}{2(t^2+9)} + \frac{1}{2\sqrt{3}} \arctan{\frac{t}{\sqrt{3}}} + C \\
= -\frac{\sqrt{x^2-9}}{2x^2} + \frac{1}{2\sqrt{3}} \arctan{\frac{\sqrt{x^2-9}}{\sqrt{3}}} + C $$
A: 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.
