<s>[INST] Reduction Integration of $I_n=\int{\frac{x^n}{\sqrt{x^2+a^2}} dx}$ How do I derive this using reduction formula?
$$I_n=\int{\frac{x^n}{\sqrt{x^2+a^2}} dx}$$
I got $\sqrt{x^2+a^2}(x^n) - \frac{1}{n-1}(x^{n-1})\sqrt{x^2+a^2} + \frac{1}{n-1}(I_n) $ . I'm not too sure if this is correct. I would appreciate any form of help or correction to my answer. Thank you.
 [/INST] For starters, I would get rid of the $a$ parameter:
$$ I_n = a^n \int \frac{x^n\,dx}{\sqrt{x^2+1}} = a^n J_n \tag{1} $$
then apply integration by parts:
$$ J_n = \int x^{n-1}\frac{x\,dx}{\sqrt{1+x^2}} = x^{n-1}\sqrt{1+x^2}-(n-1)\int x^{n-2}\sqrt{1+x^2}\,dx\tag{2}$$
and notice that $\sqrt{1+x^2}=\frac{1}{\sqrt{1+x^2}}+\frac{x^2}{\sqrt{1+x^2}}$, leading to:
$$  n\,J_n = x^{n-1}\sqrt{1+x^2} - (n-1)\,J_{n-2}.\tag{3} $$
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