Question

How do you find `int 2/((x^4-1)x) dx` using partial fractions?

Answer 1

`I=-2ln|x|+1/2ln|x-1|+1/2ln|x+1|+1/2ln|x^2+1|+C`

Explanation:

`2/((x^4-1)x)=2/((x^2-1)(x^2+1)x)=2/((x-1)(x+1)(x^2+1)x)`

`2/((x-1)(x+1)(x^2+1)x)=A/x+B/(x-1)+C/(x+1)+(Dx+E)/(x^2+1)=`

`=(A(x-1)(x+1)(x^2+1))/(x(x-1)(x+1)(x^2+1))+(Bx(x+1)(x^2+1))/(x(x-1)(x+1)(x^2+1))+`

`+(Cx(x-1)(x^2+1))/(x(x-1)(x+1)(x^2+1))+((Dx+E)x(x-1)(x+1))/(x(x-1)(x+1)(x^2+1))=`

`=(A(x^4-1))/(x(x-1)(x+1)(x^2+1))+(Bx(x^3+x^2+x+1))/(x(x-1)(x+1)(x^2+1))+`

`+(Cx(x^3-x^2+x-1))/(x(x-1)(x+1)(x^2+1))+((Dx+E)(x^3-x))/(x(x-1)(x+1)(x^2+1))=`

`=(Ax^4-A)/(x(x-1)(x+1)(x^2+1))+(Bx^4+Bx^3+Bx^2+Bx)/(x(x-1)(x+1)(x^2+1))+`

`+(Cx^4-Cx^3+Cx^2-Cx)/(x(x-1)(x+1)(x^2+1))+(Dx^4+Ex^3-Dx^2-Ex)/(x(x-1)(x+1)(x^2+1))=`

`=(x^4(A+B+C+D)+x^3(B-C+E))/(x(x-1)(x+1)(x^2+1))+`

`+(x^2(B+C-D)+x(B-C-E)+(-A))/(x(x-1)(x+1)(x^2+1))`

`A+B+C+D=0`
`B-C+E=0`
`B+C-D=0`
`B-C-E=0`
`-A=2 => A=-2`

`[II-IV] => 2E=0 => E=0`

`B+C+D=2`
`B-C=0 => B=C`
`B+C-D=0`

`2C+D=2`
`2C-D=0`

`4C=2 => C=1/2 => B=1/2`

`D=2C => D=1`

`I=int 2/((x^4-1)x)dx= -2int dx/x +1/2int dx/(x-1) + 1/2int dx/(x+1) + int (xdx)/(x^2+1)`

`I=-2ln|x|+1/2ln|x-1|+1/2ln|x+1|+1/2ln|x^2+1|+C`

Note:

`int (xdx)/(x^2+1) = int (1/2(2xdx))/(x^2+1) = 1/2int (d(x^2+1))/(x^2+1) = 1/2ln|x^2+1|`