As the degree of the numerator is equal to the degree of the denominator, perform a polynomial long division
#(4x^3+2x^2+1)/(4x^3-x)=1+(2x^2+x+1)/(4x^3-x)#
#=1+(2x^2+x+1)/(x(4x^2-1))#
#=1+(2x^2+x+1)/(x(2x+1)(2x-1))#
Perform the partial fraction decomposition
#(2x^2+x+1)/(x(2x+1)(2x-1))=A/(x)+B/(2x+1)+C/(2x-1)#
#=(A(2x+1)(2x-1)+B(x)(2x-1)+C(x)(2x+1))/(x(2x+1)(2x-1))#
The denominators are the same, compare the numerators
#2x^2+x+1=A(2x+1)(2x-1)+B(x)(2x-1)+C(x)(2x+1)#
Let #x=0#, #=>#, #1=-A#, #=>#, #A=-1#
Let #x=-1/2#, #=>#, #1=B#
Let #x=1/2#, #=>#, #C=2#
Therefore,
#(4x^3+2x^2+1)/(4x^3-x)=1-1/(x)+1/(2x+1)+2/(2x-1)#
#int((4x^3+2x^2+1)dx)/(4x^3-x)=int1dx-int(1dx)/(x)+int(1dx)/(2x+1)+int(2dx)/(2x-1)#
#=x-ln(|x|)+1/2ln(|2x+1|)+ln(|2x-1|)+C#
