# How do you find the antiderivative of int 1/(x^2(1+x^2)) dx?

Oct 16, 2016

Partial fraction expansion gives you two trivial integrals:

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

#### Explanation:

Use partial fraction expansion:

$\frac{1}{{x}^{2} \left({x}^{2} + 1\right)} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C x + D}{{x}^{2} + 1}$

$1 = A \left(x \left({x}^{2} + 1\right)\right) + B \left({x}^{2} + 1\right) + \left(C x + D\right) \left({x}^{2}\right)$

Let $x = 0$:

B = 1

$1 - \left({x}^{2} + 1\right) = A \left(x \left({x}^{2} + 1\right)\right) + \left(C x + D\right) \left({x}^{2}\right)$

Let x = 1:

$- 1 = A \left(1 \left({1}^{2} + 1\right)\right) + \left(C + D\right) \left({1}^{2}\right)$

-1 = 2A + C + D

Let x = -1

$1 - \left(- {1}^{2} + 1\right) = A \left(- 1 \left(- {1}^{2} + 1\right)\right) + \left(C - 1 + D\right) \left(- {1}^{2}\right)$

-1 = -2A - C + D

D = -1

A = C = 0

Check:

$\frac{1}{x} ^ 2 - \frac{1}{{x}^{2} + 1} =$

$\frac{1}{x} ^ 2 \frac{{x}^{2} + 1}{{x}^{2} + 1} - \frac{1}{{x}^{2} + 1} \frac{{x}^{2}}{{x}^{2}} =$

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

$\frac{1}{{x}^{2} \left({x}^{2} + 1\right)}$ This checks.

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

Oct 21, 2016

$- \frac{1}{x} - \arctan x + C$

#### Explanation:

This method avoids partial fractions and uses a trig substitution.

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

Let $x = \tan \theta$. This implies that $\mathrm{dx} = {\sec}^{2} \theta d \theta$. Thus:

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

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

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

Note that ${\cot}^{2} \theta = {\csc}^{2} \theta - 1$:

$I = \int {\csc}^{2} \theta d \theta - \int d \theta$

These are common integrals:

$I = - \cot \theta - \theta + C$

Note that $\tan \theta = x$, so $\cot \theta = \frac{1}{x}$ and $\theta = \arctan x$.

$I = - \frac{1}{x} - \arctan x + C$