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.
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).
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)
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#
