Question

`int (x^2-1)/(x^4+x^2+1) dx`?

Answer 1

`1/2Ln((x^2-x+1)/(x^2+x+1))+C`

Explanation:

`int (x^2-1)/(x^4+x^2+1)*dx`

=`int (1-1/x^2)/(x^2+1+1/x^2)*dx`

=`int ((1-1/x^2)*dx)/((x+1/x)^2-1)`

After using `y=x+1/x` and `dy=(1-1/x^2)*dx` transforms, this integral became

`int dy/(y^2-1)`

=`1/2int (2dy)/((y+1)*(y-1))`

=`1/2int dy/(y-1)-1/2int dy/(y+1)`

=`1/2Ln(y-1)-1/2Ln(y+1)+C`

=`1/2Ln((y-1)/(y+1))+C`

=`1/2Ln((x+1/x-1)/(x+1/x+1))+C`

=`1/2Ln((x^2-x+1)/(x^2+x+1))+C`

Answer 2

`int (x^2-1)/(x^4+x^2+1) dx = 1/2 ln abs(x^2-x+1)-1/2ln abs(x^2+x+1) + C`

Explanation:

Note that:

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

We find:

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

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

`color(white)((x^2-1)/(x^4+x^2+1)) = 1/2((d/(dx)(x^2-x+1))/(x^2-x+1))-1/2((d/(dx)(x^2+x+1))/(x^2+x+1))`

So:

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

`color(white)(int (x^2-1)/(x^4+x^2+1) dx) = 1/2 ln abs(x^2-x+1)-1/2ln abs(x^2+x+1) + C`