# Math Help - Stuck at reduction formula of integrating (x^2+a^2)^(-n)

1. ## Stuck at reduction formula of integrating (x^2+a^2)^(-n)

Given that $I_n=\int\frac{1}{(x^2+a^2)^n} dx$, where a and n are constants, $a\not=0$, show that, for n $\not= 0$, $2na^2I_{n+1}=(2n-1)I_n+\frac{1}{(1+a^2)^n}$
Help

2. Originally Posted by ssadi
Given that $I_n=\int\frac{1}{(x^2+a^2)^n} dx$, where a and n are constants, $a\not=0$, show that, for n $\not= 0$, $2na^2I_{n+1}=(2n-1)I_n+\frac{1}{(1+a^2)^n}$
Help
Well, here's something that's close

If $I_n = \int \frac{dx}{(x^2+a^2)^n}$ then integration by parts
$u = \frac{1}{(x^2+a^2)^n},\;\;dv = dx$ so

$du = \frac{-2nx}{(x^2+a^2)^{n+1}},\;\;\;u = x$
so

$I_n = \frac{x}{(x^2+a^2)^n} + 2n \int \frac{x^2}{(x^2+a^2)^{n+1}}\,dx = \frac{x}{(x^2+a^2)^n} + 2n \int \frac{x^2+a^2 - a^2}{(x^2+a^2)^{n+1}}\,dx$
so
$I_n = \frac{x}{(x^2+a^2)^n} +2n\left(I_n - a^2 I_{n+1} \right)$

from which we obtain

$2na^2I_{n+1}=(2n-1)I_n+\frac{x}{(x^2+a^2)^n}$