I've completed the squares in order to get a fraction in the integrand of the form $\frac {1}{\sqrt{a^2-x^2}}$ that can be easily substituted by a trigonometric function (drawing the respective triangle...).
Doing the last step in the original function I got: $\int \frac {1}{\sqrt {9-(x+2)^2}}dx$ And substituting $\sqrt{9-(x+2)^2}=3\sin q$ and $dx=-3\sin q$, I got
$\int \frac{-3\sin q}{(3\sin q)^5}dq= -\int \frac {1}{\sin^4(q)}dq = -\int \csc^4(q) .$ So the question is how do I suppose to integrate $\csc^4 (q)$?
English is not my first language so I apologize for any mistakes.