Question

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` ?

Answer 1

`int1/(x^5+4x+2)dx=sum_(i=1)^5A_iln|x-omega_i|+C`

where `A_i = prod_(j!=i)1/(omega_i-omega_j)`

Explanation:

As `x^5+4x+2` is a fifth degree polynomial with the five zeros `omega_1, omega_2,...,omega_5`, the fundamental theorem of algebra gives us that

`x^5+4x+2 = c(x-omega_1)(x-omega_2)...(x-omega_5)`

As `x^5+4x+2` is monic (has a leading coefficient of `1`), we know that `c=1`. Thus `x^5+4x+2=(x-omega_1)(x-omega_2)...(x-omega_5)`

With that, we can use partial fraction decomposition to write `1/(x^5+4x+2) = 1/((x-omega_1)(x-omega_2)...(x-omega_5))` as a sum of rational expressions with linear denominators.

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First, set

`1/((x-omega_1)(x-omega_2)...(x-omega_5)) =sum_(i=1)^5A_i/(x-omega_i)`

where `A_1, A_2, ..., 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(x-omega_i)`, 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, ..., omega_5` being distinct, but can be modified to handle cases with multiplicities greater than `1`. See the Heaviside cover up method for details).

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Note that if we multiply each side by `(x-omega_1)`, we get

`1/((x-omega_2)...(x-omega_5)) = A_1 + sum_(i=2)^5(A_i(x-omega_1))/(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 = 1/((omega_1-omega_2)(omega_1-omega_3)(omega_1-omega_4)(omega_1-omega_5))`

In fact, if we repeat this process of multiplying by `(x-omega_i)` and then setting `x=omega_i`, we will find that

`A_2 = 1/((omega_2-omega_1)(omega_2-omega_3)(omega_2-omega_4)(omega_2-omega_5))`

`A_3 = 1/((omega_3-omega_1)(omega_3-omega_2)(omega_3-omega_4)(omega_3-omega_5))`

`A_4 = 1/((omega_4-omega_1)(omega_4-omega_2)(omega_4-omega_3)(omega_4-omega_5))`

`A_5 = 1/((omega_5-omega_1)(omega_5-omega_2)(omega_5-omega_3)(omega_5-omega_4))`

or, more concisely, `A_i = prod_(j!=i)1/(omega_i-omega_j)`

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

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Now we may use the fact that `int1/(x-a)dx = ln|x-a|+C` to solve our integral.

`int1/(x^5+4x+2)dx = int(sum_(i=1)^5A_i/(x-omega_i))dx`

`=sum_(i=1)^5intA_i/(x-omega_i)dx`

`=sum_(i=1)^5A_iint1/(x-omega_i)dx`

`=sum_(i=1)^5A_iln|x-omega_i|+C`