# How do you use partial fraction decomposition to decompose the fraction to integrate (x^5 + 1)/(x^6 - x^4)?

Aug 1, 2015

$\frac{{x}^{5} + 1}{{x}^{6} - {x}^{4}} = - \frac{1}{x} ^ 2 - \frac{1}{x} ^ 4 + \frac{1}{x - 1}$

#### Explanation:

Factor the denominator:

${x}^{6} - {x}^{4} - {x}^{4} \left({x}^{2} - 1\right) = {x}^{4} \left(x + 1\right) \left(x - 1\right)$

Perhaps we know that ${x}^{5} + 1$ can be factored, or perhaps we notice that $- 1$ is a zero of the numerator, so $x - 1$ is a factor.

In any case, we can write:

$\frac{{x}^{5} + 1}{{x}^{6} - {x}^{4}} = \frac{\left(x + 1\right) \left({x}^{4} - {x}^{3} + {x}^{2} - x + 1\right)}{{x}^{4} \left(x + 1\right) \left(x - 1\right)}$

So

$\frac{{x}^{5} + 1}{{x}^{6} - {x}^{4}} = \frac{{x}^{4} - {x}^{3} + {x}^{2} - x + 1}{{x}^{4} \left(x - 1\right)}$

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

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

$= \frac{{x}^{4} - {x}^{3} + {x}^{2} - x + 1}{{x}^{4} \left(x - 1\right)}$

So
$A + E = 1$
$- A + B = - 1$
$- B + C = 1$
$- C + D = - 1$
$- D = 1$

So $D = - 1$ and,working back through the list:

$C = 0 \text{ }$, $B = - 1 \text{ }$, $A = 0 \text{ }$, and $\text{ } E = 1$.

But what if? What if we didn't notice that the fraction can be reduced?

It's OK, just a bit longer.

We would get:

$\frac{{x}^{5} + 1}{{x}^{6} - {x}^{4}} = \frac{{x}^{5} + 1}{{x}^{4} \left(x + 1\right) \left(x - 1\right)}$

$\frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x} ^ 3 + \frac{D}{x} ^ 4 + \frac{E}{x - 1} + \frac{F}{x + 1} = \frac{{x}^{5} + 1}{{x}^{4} \left(x + 1\right) \left(x - 1\right)}$

In this case we get one more variable (and one more equation).

When we solve, we'll get the same $A , B , C , D , \text{and } E$ and we will get $F = 0$