Integrate $\int \frac{\sqrt{9x^2-1}}{2x}dx$

Question

What is $\int \frac{\sqrt{9x^2-1}}{2x}dx$?

I tried to form a triangle with $\cos\theta=\frac{1}{3x}$ and $\sin\theta=\frac{\sqrt{9x^2-1}}{3x}$ to use as substitution. But I can't get rid of all the $x$'s to finally integrate with respect to $\theta$.

How is this problem solved?

Answer 1

If you do $x=\frac43\sec\theta$ and $\mathrm dx=\frac43\sec\theta\tan\theta\,\mathrm d\theta$, then your integral becomes$$\int\frac{4\tan\theta}{\frac83\sec\theta}\frac43\sec\theta\tan\theta\,\mathrm d\theta=2\int\tan^2\theta\,\mathrm d\theta.$$Can you take it from here?

Answer 2

Hint Apply the subsititon $x= \dfrac{\sec \theta}{3}$ to simplify the integral .