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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.

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    http://www.wolframalpha.com/input/?i=Integrate%5B1%2F%285-x-x%5E2%29%5E%285%2F2%29%2Cx%5D2011-11-25
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    There are a few typos, so I don't know what the problem is exactly. 1) If the problem in the title is right, then the completing the square is not; 2) In the first displayed line of the main post, the exponent is not right; 3) If the completed square was right I would make the substitution $x+2=3\sin q$; There are also constant problems if the exponent is $5/2$; 4) With the suggested substitution we do get an integral close to yours, with the more familiar $\sec$. A technique like the one posted in an answer works.2011-11-25
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    @AndréNicolas. I think you are right. I should have checked to see if he completed the square properly before posting the answer. He only has to complete the square well and then he can use the answer I have posted.2011-11-25
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    @smanoos: However, your do answer the question about how to integrate $\csc^4\theta$, and that will remain useful even if the numbers change.2011-11-25

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