What is the integral of $\frac{\sqrt{x^2-49}}{x^3}$ I used trig substitution and got 
$\displaystyle \int \dfrac{7\tan \theta}{343\sec ^3\theta}d\theta$
Then simplified to sin and cos functions, using U substitution with a final answer of:
$\dfrac{-7}{3x^3}+C$
Which section did I go wrong in. Any help would be appreciate!

If you let $x=7 \sec \theta$ then you won't get $7 \tan \theta$ in the numerator since $\sec^2\theta -\tan^2\theta =1$.
When the sub is done correctly you'll get an integral of the form  $\displaystyle \frac{49}{343} \int \frac{\tan^2\theta}{\sec^2\theta}\,d\theta = \frac{1}{7} \int \sin^2\theta\ d\theta = \frac{1}{14}\int (1-\cos 2\theta)\ d\theta$ and the rest should be easy.
