# How do you integrate int 1/(x^4 - 16) using partial fractions?

Feb 1, 2017

$\int \frac{1}{{x}^{4} - 16} \mathrm{dx} = \frac{1}{32} \ln | x - 2 | - \frac{1}{32} \ln | x + 2 | - \frac{1}{16} \arctan \left(\frac{x}{2}\right) + C$

#### Explanation:

${x}^{4} - 16 = \left({x}^{2} - 4\right) \left({x}^{2} + 4\right) = \left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 4\right)$

The partial fraction decomposition will therefore be of the form $\frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C x + D}{{x}^{2} + 4}$.

$\frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C x + D}{{x}^{2} + 4} = \frac{1}{\left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 4\right)}$

$A \left(x - 2\right) \left({x}^{2} + 4\right) + B \left(x + 2\right) \left({x}^{2} + 4\right) + \left(C x + D\right) \left(x + 2\right) \left(x - 2\right) = 1$

$A \left({x}^{3} - 2 {x}^{2} + 4 x - 8\right) + B \left({x}^{3} + 2 {x}^{2} + 4 x + 8\right) + \left(C x + D\right) \left({x}^{2} - 4\right) = 1$

$A {x}^{3} - 2 A {x}^{2} + 4 A x - 8 A + B {x}^{3} + 2 B {x}^{2} + 4 B x + 8 B + C {x}^{3} + D {x}^{2} - 4 C x - 4 D = 1$

$\left(A + B + C\right) {x}^{3} + \left(2 B - 2 A + D\right) {x}^{2} + \left(4 A + 4 B - 4 C\right) x + \left(8 B - 8 A - 4 D\right) = 1$

Write a system of equations now:

{(A + B + C = 0),(2B - 2A + D = 0), (4A + 4B - 4C = 0), (8B - 8A - 4D = 1 ):}

First of all, the third equation can be simplified to $A + B - C = 0$. We can get rid of the $C$ variable by adding the first and the third equations, to get $2 A + 2 B = 0 \to A + B = 0$. This means that $B = - A$. Also, from the second equation, we deduce that $D = 2 A - 2 B$.

Substitute into the fourth equation:

$8 \left(- A\right) - 8 A - 4 \left(2 A - 2 \left(- A\right)\right) = 1$

$- 8 A - 8 A - 8 A - 8 A = 1$

$- 32 A = 1$

$A = - \frac{1}{32}$

Solving for the other variables is now relatively simple. You should get $A = - \frac{1}{32}$, $B = \frac{1}{32}$,$C = 0$ and $D = - \frac{1}{8}$ as final answers.

$\therefore$The partial fraction decomposition is $- \frac{1}{32 \left(x + 2\right)} + \frac{1}{32 \left(x - 2\right)} - \frac{1}{8 \left({x}^{2} + 4\right)}$

Let's look closely at the integration of $\int - \frac{1}{8 \left({x}^{2} + 4\right)} \mathrm{dx}$. This will be of the arctangent form, but first we need to perform a u-substitution.

Let $u = \frac{x}{2}$. Then $\mathrm{du} = \frac{1}{2} \mathrm{dx}$.

$- \frac{1}{8} \int \frac{1}{{u}^{2} + 1} \cdot \frac{1}{2} \mathrm{du}$

$- \frac{1}{16} \int \frac{1}{{u}^{2} + 1} \mathrm{du}$

This is a standard integral.

$- \frac{1}{16} \arctan u$

$- \frac{1}{16} \arctan \left(\frac{x}{2}\right)$

Now deal with the other parts of the problem. The other two terms can be integrated using $\int \frac{1}{x} \mathrm{dx} = \ln | x | + C$.

$\int \frac{1}{32 \left(x - 2\right)} \mathrm{dx} = \frac{1}{32} \ln | x - 2 |$

$\int - \frac{1}{32 \left(x + 2\right)} \mathrm{dx} = - \frac{1}{32} \ln | x + 2 |$

Therefore, $\int \frac{1}{{x}^{4} - 16} \mathrm{dx} = \frac{1}{32} \ln | x - 2 | - \frac{1}{32} \ln | x + 2 | - \frac{1}{16} \arctan \left(\frac{x}{2}\right) + C$

Hopefully this helps!