# How do you evaluate the integral int dx/(x^4+x^2)?

Jan 4, 2017

The answer is $= - \frac{1}{x} - \arctan x + C$

#### Explanation:

We need

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$

We factorise the denominator

${x}^{4} + {x}^{2} = {x}^{2} \left({x}^{2} + 1\right)$

Now we can do the decomposition into partial fractions

$\frac{1}{{x}^{4} + {x}^{2}} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C x + D}{{x}^{2} + 1}$

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

Therefore,

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

Let, $x = 0$,$\implies$,$1 = A$

Coefficients of ${x}^{2}$, $\implies$, $0 = A + D$, $\implies$, $D = - 1$

Coefficients of $x$, $\implies$, $0 = B$

Coeficients of ${x}^{3}$, $\implies$, $0 = B + C$, $\implies$, $C = 0$

So,

$\frac{1}{{x}^{4} + {x}^{2}} = \frac{1}{x} ^ 2 + \frac{0}{x} + \frac{0 x - 1}{{x}^{2} + 1}$

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

Therefore,

$\int \frac{\mathrm{dx}}{{x}^{4} + {x}^{2}} = \int \frac{\mathrm{dx}}{x} ^ 2 - \int \frac{\mathrm{dx}}{{x}^{2} + 1}$

The first integral is $\int \frac{\mathrm{dx}}{x} ^ 2 = - \frac{1}{x}$

The second integral is $\int \frac{\mathrm{dx}}{{x}^{2} + 1}$

We use a trigonometric substitution

Let $x = \tan u$, $\implies$, $\mathrm{du} = {\sec}^{2} u \mathrm{du}$

and ${x}^{2} + 1 = {\tan}^{2} u + 1 = {\sec}^{2} u$

So,

$\int \frac{\mathrm{dx}}{{x}^{2} + 1} = \int \frac{{\sec}^{2} u \mathrm{du}}{\sec} ^ 2 u = \int \mathrm{du} = u$

$= \arctan x$

Putting it alltogether,

$\int \frac{\mathrm{dx}}{{x}^{4} + {x}^{2}} = - \frac{1}{x} - \arctan x + C$