# How do you integrate 1/((1+x^2)^2)?

Oct 2, 2016

$\frac{x}{2 + 2 {x}^{2}} + \arctan \frac{x}{2} + C$

#### Explanation:

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

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

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

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

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

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 $\theta = \arctan x$.

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

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

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

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