# Evaluate the integral? : int x^2/(x^2+1)^2 dx

## (Question Restore: portions of this question have been edited or deleted!)

Jul 30, 2017

Option #1

$\int \setminus {x}^{2} / {\left({x}^{2} + 1\right)}^{2} \mathrm{dx} = \frac{1}{2} \arctan x - \frac{x}{2 \left({x}^{2} + 1\right)} + C$

#### Explanation:

Let us denote the given integral by $I$:

$I = \int \setminus {x}^{2} / {\left({x}^{2} + 1\right)}^{2} \mathrm{dx}$

We can simply the fractional integrand with a slight manipulation:

$I = \int \setminus \frac{{x}^{2} + 1 - 1}{{x}^{2} + 1} ^ 2 \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{{x}^{2} + 1}{{x}^{2} + 1} ^ 2 - \frac{1}{{x}^{2} + 1} ^ 2 \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{1}{{x}^{2} + 1} \setminus \mathrm{dx} - \int \setminus \frac{1}{{x}^{2} + 1} ^ 2 \setminus \mathrm{dx}$
$\setminus \setminus = \arctan x - \int \setminus \frac{1}{{x}^{2} + 1} ^ 2 \setminus \mathrm{dx} + C$

For this next integral we can use a substitution: Let

$\tan \theta = x \implies {x}^{2} + 1 = {\tan}^{2} \theta + 1 = {\sec}^{2} \theta$
$\frac{\mathrm{dx}}{d \theta} = {\sec}^{2} \theta$

Substituting we see that:

$\int \setminus \frac{1}{{x}^{2} + 1} ^ 2 \setminus \mathrm{dx} = \int \setminus \frac{1}{{\sec}^{2} \theta} ^ 2 \setminus {\sec}^{2} \theta \setminus d \theta$
$\text{ } = \int \setminus \frac{1}{{\sec}^{2} \theta} ^ 2 \setminus {\sec}^{2} \theta \setminus d \theta$
$\text{ } = \int \setminus \frac{1}{{\sec}^{2} \theta} \setminus d \theta$
$\text{ } = \int \setminus {\cos}^{2} \theta \setminus d \theta$
$\text{ } = \int \setminus \frac{1}{2} \left(1 + \cos 2 \theta\right) \setminus d \theta$
$\text{ } = \frac{1}{2} \setminus \int \setminus 1 + \cos 2 \theta \setminus d \theta$
$\text{ } = \frac{1}{2} \left(\theta + \frac{1}{2} \sin 2 \theta\right)$
$\text{ } = \frac{1}{2} \left(\theta + \sin \theta \cos \theta\right)$
$\text{ } = \frac{1}{2} \left(\theta + \sin \theta \cos \theta \cdot \cos \frac{\theta}{\cos} \theta\right)$
$\text{ } = \frac{1}{2} \left(\theta + \sin \frac{\theta}{\cos} \theta {\cos}^{2} \theta\right)$
$\text{ } = \frac{1}{2} \left(\theta + \tan \frac{\theta}{\sec} ^ 2 \theta\right)$
$\text{ } = \frac{1}{2} \left(\arctan x + \frac{x}{{x}^{2} + 1}\right)$

Combining this with the earlier result gives:

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