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

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

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

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

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

from which we obtain

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

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# reduction formulae for integral (x^2 a^2)^7รท2

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