How to find integrals of the form $\frac 1{(a^2+x^2)^n}$

Question

How will we find the integrals of the form $\frac 1{(a^2+x^2)^n}$?
For example if we find the integral of $\frac{1}{(x^2+1)^3}$, I cannot see any substitution here. Moreover, if we do integration by parts, it becomes more and more messy. I have heard somewhere about reduction formula, so please can someone explain the derivation of it and why does it work?

Answer 1

Let your integral be $I_n$. Then we use integral by part:

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

$dx = dv, v = x$

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

The integral on the RHS can be written as: $\int \frac{2n(x^2+a^2)}{(a^2+x^2)^{n+1}}dx - \int \frac{2na^2}{(a^2+x^2)^{n+1}}dx = 2nI_n-2na^2I_{n+1} $

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

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

From this recursive formula, you can calculate $I_n$ (It's gonna be very lengthy, so I won't write it explicitly here)

Answer 2

Suppose $a\not=0,$ then we know that$$\int_0^y\dfrac{1}{a^2+x^2}dx=\arctan \dfrac{y}{a}$$ Then you can use differentiation under integration sign with respect to $a.$