Question

How do you integrate $\frac{1}{(1+x^2{)}^2}$?

Answer 1

$\frac{x}{2+2x^2}+\frac{arc\tan x}{2}+C$

Explanation:

$I=\int \frac{dx}{(1+x^2{)}^2}$

We will use the substitution $x=\tan \theta$, implying that $dx={\sec }^2\theta d\theta$:

$I=\int \frac{{\sec }^2\theta d\theta }{(1+{\tan }^2\theta {)}^2}$

Note that $1+{\tan }^2={\sec }^2\theta$:

$I=\int \frac{{\sec }^2\theta d\theta }{{\sec }^4\theta }=\int \frac{d\theta }{{\sec }^2\theta }=\int {\cos }^2\theta d\theta$

Recall that $\cos 2\theta =2{\cos }^2\theta -1$, so ${\cos }^2\theta =\frac{1}{2}\cos 2\theta +\frac{1}{2}$.

$I=\frac{1}{2}\int \cos 2\theta d\theta +\int \frac{1}{2}d\theta$

The first integral can be found with substitution (try $u=2\theta$).

$I=\frac{1}{4}\sin 2\theta +\frac{1}{2}\theta +C$

From $x=\tan \theta$ we see that $\theta =arc\tan x$.

Furthermore, we see that $\frac{1}{4}\sin 2\theta =\frac{1}{4}\left(2\sin \theta \cos \theta \right)=\frac{1}{2}\sin \theta \cos \theta$

Also, since $\tan \theta =x$, we can draw a right triangle with the side opposite $\theta$ being $x$ the adjacent side being $1$, and the hypotenuse being $\sqrt{1+x^2}$. Thus,

$\sin \theta =\frac{x}{\sqrt{1+x^2}}$ and $\cos \theta =\frac{1}{\sqrt{1+x^2}}$:

$I=\frac{1}{2}\sin \theta \cos \theta +\frac{1}{2}arc\tan x+C$

$I=\frac{1}{2}\left(\frac{x}{\sqrt{1+x^2}}\right)\left(\frac{1}{\sqrt{1+x^2}}\right)+\frac{arc\tan x}{2}+C$

$I=\frac{x}{2(1+x^2)}+\frac{arc\tan x}{2}+C$