How do you use partial fractions to find the integral $int x/(16x^4-1)dx$ ?

Question

2809 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-24T09:35:40

The answer is $=1/16ln(|2x+1|)+1/16ln(|2x-1|)-1/16ln(|4x^2+1|)+C$

We need

$intdx/x=ln(|x|)+C$

Let's factorise the denominator

$16x^4-1=(4x^2-1)(4x^2+1)=(2x+1)(2x-1)(4x^2+1)$

We can perform the decomposition into partial fractions

$x/(16x^4-1)=x/((2x+1)(2x-1)(4x^2+1))$

$=A/(2x+1)+B/(2x-1)+(Cx+D)/(4x^2+1)$

$=(A(2x-1)(4x^2+1)+B(2x+1)(4x^2+1)+(Cx+D)(2x+1)(2x-1))/((2x+1)(2x-1)(4x^2+))$

The denominators are the same, we compare the numerators

$x=A(2x-1)(4x^2+1)+B(2x+1)(4x^2+1)+(Cx+D)(2x+1)(2x-1)$

Let $x=-1/2$ , $=>$ , $-1/2=-4A$ , $=>$ , $A=1/8$

Let $x=1/2$ , $=>$ , $1/2=4B$ , $=>$ , $B=1/8$

Let $x=0$ , $=>$ , $0=-A+B-D$ , $=>$ , $D=0$

Coefficients of $x^3$ ,

$0=8A+8B+4C$ , $=>$ , $4C=-2$ , $C=-1/2$

Therefore,

$x/(16x^4-1)=(1/8)/(2x+1)+(1/8)/(2x-1)+(-1/2x)/(4x^2+1)$

$int(xdx)/(16x^4-1)=int(1/8dx)/(2x+1)+int(1/8dx)/(2x-1)+int(-1/2xdx)/(4x^2+1)$

$=1/16ln(|2x+1|)+1/16ln(|2x-1|)-1/16ln(|4x^2+1|)+C$

How do you use partial fractions to find the integral int x/(16x^4-1)dxint x/(16x^4-1)dx?