I was working on finding the reduction formula for :

$\int \frac{dx}{(x^2+a^2)^n}$ By using integration by parts formula ( $\int f(x) g(x) dx = f(x) \int g(x)dx -\int (f'(x)\int g(x))$ considering 1.dx as second function :

Let $I_n$ = $\int \frac{dx}{(x^2+a^2)^n} = \frac{dx}{(x^2+a^2)^n}.\int dx -\int \{\frac{dx}{(x^2+a^2)^n}.\int \}dx$

=$\frac{dx}{(x^2+a^2)^n}-\int (x^2+a^2)^{n-1}.2x^2dx$

=$\frac{dx}{(x^2+a^2)^n} +2n\int \frac{x^2}{(x^2+a^2)^{n+1}}dx$

=$\frac{dx}{(x^2+a^2)^n} + \frac{2n}{(x^2+a^2)}\int \frac{dx}{(x^2+a^2)^{n-1}}-\frac{dx}{(x^2+a^2)^n}$

Therefore , $I = \frac{x}{(x^2+a^2)^n}+2nI_n-2n^2I_{n+1}$

Please guide further how to proceed and conclude..Thanks


If $$I_n=\int \frac{dx}{(x^2+a^2)^n}$$ then setting $$u=\frac{1}{(x^2+a^2)^n},~~dv=dx$$ leads us to have $$du=\frac{-2nxdx}{(x^2+a^2)^{n+1}},~~v=x$$ and the method of integrating by parts gives us $$I_n=\frac{x}{(x^2+a^2)^n}+2n\int\frac{x^2dx}{(x^2+a^2)^{n+1}}=...=\frac{x}{(x^2+a^2)^n}+2nI_n-2na^2I_{n+1}$$ Now we see the integrating of $I_{n+1}$ will be dependent to integrating of $I_n$ and....

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  • $\begingroup$ Nice work, Babak! $\endgroup$ – amWhy May 30 '13 at 0:19
  • $\begingroup$ No :) The sun is coming up here. :-) $\endgroup$ – mrs May 30 '13 at 0:42
  • $\begingroup$ Ohhh, okay ;-) Enjoy the sun! - We've been a bit rainy, lately :-( $\endgroup$ – amWhy May 30 '13 at 0:43
  • $\begingroup$ I love rain Amy, it makes me the feeling of a romantic time. I hope I am good today at the site. Let me go for having cereal with milk. :P $\endgroup$ – mrs May 30 '13 at 0:52
  • $\begingroup$ It is a good answer +1 $\endgroup$ – Adi Dani May 30 '13 at 20:00

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