<s>[INST] Trig substitution integration of $\int1/(x^2\sqrt{x^2 - 9}) dx$ - stuck on a problem I am getting stuck on this trig substitution problem.
$$\int\frac1{x^2\sqrt{x^2 - 9}}~\mathrm dx.$$
$$x = 3 \sec\theta,\qquad\theta = \sec^{-1} \sqrt{\frac{x^2}{9}},\qquad\mathrm dx = \sec\theta\tan\theta\ \mathrm d\theta$$
I can get to here, but I don't know how to finish it (perhaps I have made a mistake before this point?)
$$\int\frac{3\sec\theta\tan\theta}{9\sec^2\theta(3\sec\theta -3)}~\mathrm d\theta.$$
If anyone could help from here, I'd appreciate it.
Thanks.
 [/INST] Using your substitution $x=3\sec \theta $ and cancelling $\sec \theta$ in the numerator and denominator, I got 
$$I=\int \frac{1}{x^{2}\sqrt{x^{2}-9}}\ \textrm{d}x=\int \frac{ \tan \theta }{3\  \sec
\theta \ \sqrt{9\sec ^{2}\theta -9}}\ \textrm{d}\theta=\int \frac{\tan \theta }{9\  \sec 
\theta \  \sqrt{\sec ^{2}\theta -1}}\ \textrm{d}\theta.$$ 
It is easy to see that
$$\frac{\tan \theta }{ \sec \theta  \sqrt{\sec ^{2}\theta -1}}=
\cos \theta .$$ 
So
$$I=\int \frac{1}{9}\cos \theta \ \textrm{d}\theta = \dots .$$ 
Added. Just to confirm Christian Blatter's evaluation. 
$$\begin{eqnarray*}
I &=&\int \frac{1}{9}\cos \theta \,d\theta =\frac{1}{9}\sin \theta +C \\
&=&\frac{1}{9}\sqrt{1-\cos ^{2}\theta }+C=\frac{1}{9}\sqrt{1-\frac{1}{\sec
^{2}\theta }} +C\\
&=&\frac{1}{9}\sqrt{1-\frac{9}{x^{2}}}+C=\frac{\sqrt{x^{2}-9}}{9x}+C.
\end{eqnarray*}$$
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