TITLE: Trig substitution integration of $\int1/(x^2\sqrt{x^2 - 9}) dx$ - stuck on a problem
QUESTION [8 upvotes]: 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.

REPLY [4 votes]: I'd suggest to use the substitution $x:=3\cosh t$ instead. This leads to
$$\eqalign{I&=\int{1\over 9\cosh^2 t\ 3\sinh t}\ 3\sinh t\ dt=\int{1\over 9\cosh^2 t}\ dt\cr &={1\over9}\tanh t+C={\sqrt{\cosh^2 t -1}\over9\cosh t}+C={\sqrt{x^2-9}\over 9 x}+C\ .\cr}$$