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Ive been struggling to find the integral $\frac{1}{(x^2+2)^3}$ by using the integral $I_n=\frac{1}{(x^2+1)^n}$. (assume I know how to solve $I_n$ by a recursive way. Ive tried to make it to the form of In but without any success, any clue?

Thanks

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1 Answer 1

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Hint :

You can notice that : $$\frac{1}{(x^2+2)^3}=\frac{1}{(2(\frac{x^2}{2}+1))^3}=\frac{1}{8((\frac{x}{\sqrt{2}})^2+1)^3}=\frac{\frac{1}{8}}{((\frac{x}{\sqrt{2}})^2+1)^3}$$ Now you should be able to use a result about $I_n$.

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  • $\begingroup$ yeah I figured that out few moments after I published the question. I feel stupid about this question. thanks anyway! $\endgroup$
    – Ami Gold
    May 27, 2016 at 8:26

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