How do you integrate $(4x^3+2x^2+1)/(4x^3-x)$ using partial fractions?

Question

3760 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-20T14:47:20

The answer is $=x-ln(|x|)+1/2ln(|2x+1|)+ln(|2x-1|)+C$

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,

$(4x^3+2x^2+1)/(4x^3-x)=1-1/(x)+1/(2x+1)+2/(2x-1)$

$int((4x^3+2x^2+1)dx)/(4x^3-x)=int1dx-int(1dx)/(x)+int(1dx)/(2x+1)+int(2dx)/(2x-1)$

$=x-ln(|x|)+1/2ln(|2x+1|)+ln(|2x-1|)+C$

How do you integrate (4x^3+2x^2+1)/(4x^3-x)(4x^3+2x^2+1)/(4x^3-x) using partial fractions?