Partial fractions expansion problem $\frac{x^3-1}{4x^3-x}$

Question

I want to calculate integral of the fraction, but first how to find the partial fraction expansion of $\frac{x^3-1}{4x^3-x}$. How to expand denominator? I am a bit lost here.

Answer 1

Divide the numerator by denominator first..

$$x^3-1=\frac 1 4 (4x^3-x)-\frac 1 4 (x-4)\tag{division algorithm}$$

$$\frac{x^3-1}{4x^3-x}=\frac 1 4 \frac{4x^3-x}{4x^3-x}-\frac 1 4\frac{x-4}{4x^3-x} $$

$$\frac{x^3-1}{4x^3-x}=\frac 1 4 -\frac 1 4\frac{x-4}{4x^3-x} $$ $$\int\frac{x^3-1}{4x^3-x}dx=\int\frac 1 4 dx-\frac 1 4\int\frac{x-4}{4x^3-x} dx$$

after this easy know...!!!! using partial fraction method, we will have

$$\dfrac{x-4}{4x^3-x}=\frac{x-4}{x(2x-1)(2x+1)}=\frac A{x}+\frac B{2x-1}+\frac C {2x+1} $$

Find $A,B,C$ then solve then...

Answer 2

$4x^3- x = x(2x+1)(2x-1).$ So write out the denominator like that and see if you can take it from there?