$\int\frac1{(1+x^2)^3}\,dx$ without Hermite

Question

$$\int\frac1{(1+x^2)^3}\,dx$$ I know this can be integrated with Hermite polynomials, but I would like to find another method, possibly substitution. I think that the right one is $x=\arctan y$. The problem is that I don't know how to manipulate the terms to follow the right steps, and the right order to simplify the factors, could you help me? Thanks.

Answer 1

Hint:

use (two times) the reduction formula: $$ \int \frac{dx}{(x^2+a^2)^n}=\frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}\int\frac{dx}{(x^2+a^2)^{n-1}} $$

(with $a=1$ and $n=3$)

Answer 2

We may try $x=\tan (u)$ and $dx = \sec^2 (u) \,du$

$$\int \frac{\sec^2 (u) \, du}{\sec^6( u)} = \int \cos^4 (u) \, du$$

Now try integration by parts

$$I=\sin (u) \cos^3 (u) - 3\int \sin^2( u) \cos^2 (u)\, du$$

The last integral $\sin^2(u) \cos^2(u)$ can be written in terms of $\frac{\sin^2(2u)}{4}$ and then perhaps use $\sin^2(2u) = \frac{1-\cos(4u)}{2}$. This leads to

$$I=\sin (u) \cos^3 (u) - \frac{3}{8}\int (1-\cos(4u))\, du\\ = \sin (u) \cos^3 (u) - \frac{3}{8}u + \frac{3}{32}\sin(4u)+c$$

Answer 3

Consider the alternatively simpler integral,

$$I_a(x)=\int\frac1{a+x^2}{\rm~d}x=\frac1{\sqrt a}\arctan\frac x{\sqrt a}+C$$

Take 2 derivatives w.r.t. $a$ and we have

$$\int\frac2{(a+x^2)^3}{\rm~d}x=\frac{\partial^2}{\partial a^2}\frac1{\sqrt a}\arctan\frac x{\sqrt a}+C$$