# How do I find the partial fraction decomposition of (x^4+1)/(x^5+4x^3) ?

##### 1 Answer
Aug 13, 2014

First we will factor the denominator as much as possible:

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

And now, we will choose the factors to write:

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

Note that since there was a lone power of $x$, (which was ${x}^{3}$) I wrote out successive powers of $x$, starting at $x$ to the first, and ending at ${x}^{3}$. There was also a quadratic term, ${x}^{2} + 4$, which couldn't be factored - so for that one, we used $D x + E$ in the numerator.

The next step is to multiply both sides of the equation by ${x}^{3} \cdot \left({x}^{2} + 4\right)$ and cancel off what we can:

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

And now, we will distribute and simplify everything:

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

We can solve for each constant now, by using the technique of grouping. The first step is to rearrange everything in successive powers of $x$:

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

And now, we will factor out the constant terms:

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

The next step is to create a system of equations using the coefficients of $x$ on the left side that correspond to the coefficients of $x$ on the right side. What do I mean? Well, we can see that there is a term $1 \cdot {x}^{4}$ on the left side, but there is also a $\left(A + D\right) \cdot {x}^{4}$ term on the right side.

This implies that $A + D = 1$. We will continue in this manner, building a system using all the coefficients:

$A + D = 1$
$B + E = 0$
$4 A + C = 0$
$4 B = 0$
$4 C = 1$

Immediately from the last two equations, we can conclude that $B = 0$ and $C = \frac{1}{4}$.

From this it follows that since $B + E = 0$, $E$ must also equal $0$. And since $4 A + C = 0$, $A$ must equal $- \frac{1}{16}$.

Then, after plugging $A$ into the last equation $A + D = 1$, and solving for $D$, we obtain $D = \frac{17}{16}$.

Now all that's left is to plug these coefficient values into our expanded expression:

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

$\frac{{x}^{4} + 1}{{x}^{3} \left({x}^{2} + 4\right)} = \frac{1}{4 {x}^{3}} + \frac{17 x}{16 \left({x}^{2} + 4\right)} - \frac{1}{16 x}$

And there we have it. Remember, successfully expanding with partial fractions is all about choosing the correct factors, and from there it's just a lot of algebra. If you are familiar with the grouping technique, then you shouldn't have any trouble solving for the coefficients.