Trig Substitution Indefinite Integral help

Question

I'm having trouble with this trig substitution problem. Please help me out.

$$\int{\frac{ \sqrt{9x^2-289}}{x}}\;dx$$

Answer 1

Hint: $$\int{\frac{ \sqrt{9x^2-289}}{x}}\;dx = \int\frac{\sqrt{(3x)^2 - (17)^2}}{x}$$

Now put $$3x = 17\sec\theta \implies x = \frac{17}3 \sec \theta \implies dx = \frac {17}{3} \tan \theta \sec \theta\, d\theta,\; \\ \theta = \sec^{-1}\left(\frac {3x}{17}\right)\;\text{ and }\;\sqrt{9x^2 - 289} = 17\sqrt{\sec^2\theta - 1} = 17\sqrt{\tan^2\theta} = 17\tan \theta$$

If you proceed with the substitution correctly, you'll arrive at $$17 \int \tan^2 \theta \,d\theta = 17\int(\sec^2\theta - 1)\,d\theta = 17\tan \theta - 17\theta + C$$

Now, we determined above both $17 \tan\theta$ and $\theta$, each in terms of $x$, based on your original substitution. Using these "back-substitutions", you are good to go!