How can I finish integrating $\int {\sqrt{x^2-49} \over x} $ using trig substitution? $$\int {\sqrt{x^2-49} \over x}\,dx $$
$$ x = 7\sec\theta$$
$$ dx = 7\tan\theta \sec\theta \,d\theta$$
$$\int {\sqrt{7^2\sec^2\theta - 7^2} \over 7\sec\theta}\left(7\tan\theta \sec\theta \,d\theta\right) = \int \sqrt{7^2\sec^2\theta - 49} \left(\tan\theta d\theta\right)$$
$$ \int\sqrt{7^2(\sec^2\theta - 1)} (\tan\theta \,d\theta) = 7\int\sqrt{\sec^2\theta - 1} (\tan\theta \,d\theta)$$
$$ 7\int \tan^2\theta \,d\theta = 7\int \sec^2\theta - 1 \,d\theta  $$
$$ 7\int \sec^2\theta - 7\int d\theta $$ 
$$ 7\tan\theta - 7\theta + C  = 7(\tan\theta - \theta) + C$$
This makes: $$ \theta = \sec^{-1}\left(x \over 7\right)$$
And plugging back in to the indefinite integral:
$$ 7\left(\left(\sqrt{x^2-49} \over 7 \right) - \sec^{-1}\left(x \over 7 \right)\right) + C $$
My question really is, how can I evaluate $\sec^{-1}\left(x \over 7 \right)$ ?

This is the correct answer. Unless the original problem asked for a definite integral, in which case all you would need to do is plug in your bounds. 
