# If the zeros of x^5+4x+2 are omega_1, omega_2,.., omega_5, then what is int 1/(x^5+4x+2) dx ?

Jun 19, 2016

$\int \frac{1}{{x}^{5} + 4 x + 2} \mathrm{dx} = {\sum}_{i = 1}^{5} {A}_{i} \ln | x - {\omega}_{i} | + C$

where ${A}_{i} = {\prod}_{j \ne i} \frac{1}{{\omega}_{i} - {\omega}_{j}}$

#### Explanation:

As ${x}^{5} + 4 x + 2$ is a fifth degree polynomial with the five zeros ${\omega}_{1} , {\omega}_{2} , \ldots , {\omega}_{5}$, the fundamental theorem of algebra gives us that

${x}^{5} + 4 x + 2 = c \left(x - {\omega}_{1}\right) \left(x - {\omega}_{2}\right) \ldots \left(x - {\omega}_{5}\right)$

As ${x}^{5} + 4 x + 2$ is monic (has a leading coefficient of $1$), we know that $c = 1$. Thus ${x}^{5} + 4 x + 2 = \left(x - {\omega}_{1}\right) \left(x - {\omega}_{2}\right) \ldots \left(x - {\omega}_{5}\right)$

With that, we can use partial fraction decomposition to write $\frac{1}{{x}^{5} + 4 x + 2} = \frac{1}{\left(x - {\omega}_{1}\right) \left(x - {\omega}_{2}\right) \ldots \left(x - {\omega}_{5}\right)}$ as a sum of rational expressions with linear denominators.

First, set

$\frac{1}{\left(x - {\omega}_{1}\right) \left(x - {\omega}_{2}\right) \ldots \left(x - {\omega}_{5}\right)} = {\sum}_{i = 1}^{5} {A}_{i} / \left(x - {\omega}_{i}\right)$

where ${A}_{1} , {A}_{2} , \ldots , {A}_{5}$ are the (currently unknown) constants which would make the above equation true.

While we could now multiply each side by ${\prod}_{i = 1}^{5} \left(x - {\omega}_{i}\right)$, simplify, equate corresponding coefficients, and then solve the resulting system of equations with five variables, instead we will use a more convenient method (note that the following relies on ${\omega}_{1} , {\omega}_{2} , \ldots , {\omega}_{5}$ being distinct, but can be modified to handle cases with multiplicities greater than $1$. See the Heaviside cover up method for details).

Note that if we multiply each side by $\left(x - {\omega}_{1}\right)$, we get

$\frac{1}{\left(x - {\omega}_{2}\right) \ldots \left(x - {\omega}_{5}\right)} = {A}_{1} + {\sum}_{i = 2}^{5} \frac{{A}_{i} \left(x - {\omega}_{1}\right)}{x - {\omega}_{i}}$

If we set $x = {\omega}_{1}$, then every term on the right hand side except for ${A}_{1}$ becomes $0$, and so we find that

${A}_{1} = \frac{1}{\left({\omega}_{1} - {\omega}_{2}\right) \left({\omega}_{1} - {\omega}_{3}\right) \left({\omega}_{1} - {\omega}_{4}\right) \left({\omega}_{1} - {\omega}_{5}\right)}$

In fact, if we repeat this process of multiplying by $\left(x - {\omega}_{i}\right)$ and then setting $x = {\omega}_{i}$, we will find that

${A}_{2} = \frac{1}{\left({\omega}_{2} - {\omega}_{1}\right) \left({\omega}_{2} - {\omega}_{3}\right) \left({\omega}_{2} - {\omega}_{4}\right) \left({\omega}_{2} - {\omega}_{5}\right)}$

${A}_{3} = \frac{1}{\left({\omega}_{3} - {\omega}_{1}\right) \left({\omega}_{3} - {\omega}_{2}\right) \left({\omega}_{3} - {\omega}_{4}\right) \left({\omega}_{3} - {\omega}_{5}\right)}$

${A}_{4} = \frac{1}{\left({\omega}_{4} - {\omega}_{1}\right) \left({\omega}_{4} - {\omega}_{2}\right) \left({\omega}_{4} - {\omega}_{3}\right) \left({\omega}_{4} - {\omega}_{5}\right)}$

${A}_{5} = \frac{1}{\left({\omega}_{5} - {\omega}_{1}\right) \left({\omega}_{5} - {\omega}_{2}\right) \left({\omega}_{5} - {\omega}_{3}\right) \left({\omega}_{5} - {\omega}_{4}\right)}$

or, more concisely, ${A}_{i} = {\prod}_{j \ne i} \frac{1}{{\omega}_{i} - {\omega}_{j}}$

(This process shows the reasoning behind the Heaviside cover up method for solving partial fractions)

Now we may use the fact that $\int \frac{1}{x - a} \mathrm{dx} = \ln | x - a | + C$ to solve our integral.

$\int \frac{1}{{x}^{5} + 4 x + 2} \mathrm{dx} = \int \left({\sum}_{i = 1}^{5} {A}_{i} / \left(x - {\omega}_{i}\right)\right) \mathrm{dx}$

$= {\sum}_{i = 1}^{5} \int {A}_{i} / \left(x - {\omega}_{i}\right) \mathrm{dx}$

$= {\sum}_{i = 1}^{5} {A}_{i} \int \frac{1}{x - {\omega}_{i}} \mathrm{dx}$

$= {\sum}_{i = 1}^{5} {A}_{i} \ln | x - {\omega}_{i} | + C$