If $$I_n = \int{(\frac{1}{a^2+x^2})^{n}}dx$$
Prove that:$$I_n = \frac{x}{2a^2(n-1)(a^2+x^2)^{(n-1)}}+\frac{2n-3}{2(n-1)a^2}I_{n-1}$$
I used Ibp but couldn't get such a relation. Please help me. Also, please do not use induction.
2026-04-13 13:36:22.1776087382
Reduction formula doubt.
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Hint: By parts, setting $\,u= \dfrac1{(a^2+x^2)^{n-1}}$, $\,\mathrm d\mkern1mu v=\mathrm d\mkern1mu x$, whence $$\mathrm d\mkern1mu u=\frac{-2(n-1)x}{(a^2+x^2)^n}\,\mathrm d\mkern1mu x,\enspace v=x $$ One obtains then $$I_{n-1}=\dfrac x{(a^2+x^2)^{n-1}}+2(n-1)\int\frac{x^2}{(a^2+x^2)^n}\,\mathrm d\mkern1mu x$$ Note that, writing $\,x^2=a^2+x^2 -a^2$, the integral is equal to $\,I_{n-1}-a^2I_n$.