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Prove $$\int \frac{dx}{(1+x^2)^n} = \frac{1}{2n-2}\frac{x}{(x^2+1)^{n-1}}+\frac{2n-3}{2n-2}\int\frac{1}{(x^2+1)^{n-1}}dx$$ by writing $$\int \frac{dx}{(1+x^2)^n}=\int \frac{dx}{(1+x^2)^{n-1}} - \int\frac{x^2}{(1+x^2)^n}dx$$ and working on the last integral. (Another possibility is to use the substitution $x=\tan u$.)

I was able to do it the first way. Now I'm trying to do it with the substitution $x=\tan u$.

$$\int\frac{\sec^2 u}{(\sec^2 u)^n}du=\int\frac{du}{(\sec^2 u)^{n-1}}.$$ Now I was fiddling around with this, thinking of trying to integrate by parts using $dv = du$ and $u$ being the rest. But I'm just kind of wandering with this one. Is this how you would proceed?

clathratus
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Eric Auld
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    $\int\frac{du}{(\sec^2 u)^{n-1}}=\int (\cos^2 u)^{n-1}du$ – David H Jan 26 '14 at 07:27
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    Taking into account what David H wrote, have a look at http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions#Integrands_involving_only_cosine. You will find the recurrence relation you look for. – Claude Leibovici Jan 26 '14 at 07:34

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