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!

Alternately, letting $x=7\cosh t$, and using the fact that $\cosh^2t-\sinh^2t=1,~\cosh't=\sinh t$, and $\sinh't=\cosh t$, we ultimately arrive at $$\displaystyle\int\frac{\sinh^2t}{\cosh^3t}dt=\int\frac{\sinh^2t~\cosh t}{\cosh^4t}dt=\int\frac{\sinh^2t\cdot d\big(\sinh t\big)}{\big(1+\sinh^2t\big)^2}=\int\frac{u^2}{\big(1+u^2\big)^2}du,~$$ which is trivial.
